Definitions and Goal
Definition: A field $K$ with the property $K=K^2$ is called square-root-closed (src).
Definition: A field $K$ is called odd, if every irreducible polynomial $f\in K[X]$ has odd degree.
A field $K$ of characteristic $p\neq 2$ is odd if and only every separable, irreducible polynomial $f\in K[X]$ has odd degree.
Every separably closed field of characteristic $p\neq 2$ is odd.
A field $K$ of characteristic $p\neq 2$ is odd if and only if every finite extension has odd degree.
Every algebraic extension of an odd field of characteristic $p\neq 2$ is odd.
Every odd field is src.
Question: Is every src-field $K$ of characteristic $p\neq 2$ odd?
Partial answer given in this post: the answer is Yes for the following classes of fields:
Class 1: $K$ an algebraic extension of a finite field.
Class 2: $K$ an algebraic extension of the rationals, that carries a valuation $v$ such that the residue field $k(v)$ has characteristic $p\neq 2$ and $(K,v)$ is henselian.
The method provided here to prove the answer is Yes for the fields in class 2 actually can be applied recursively thus creating infinitely many classes of fields for which the answer is Yes. Precisely the following result is proved:
Theorem: let $K$ be an src-field that carries a valuation $v$ such that $(K,v)$ is henselian, the characteristic of $k(v)$ is $\neq 2$ and $k(v)$ ia an odd field. Then $K$ is odd.
Proof of the assertion for fields in class 1
The square-root-closure of a finite field $\mathbb{F}_q$ with $q=p^n$ elements, $p\neq 2$.
(A) In $\mathbb{F}_q$ one has $-1\neq 1$ and $(-1)^2=1$. Thus $\mathbb{F}_q\neq\mathbb{F}_q^2$.
(B) In a fixed algebraic closure of $\mathbb{F}_q$ there is exactly one extension $F/\mathbb{F}_q$ of degree $2$, namely $\mathbb{F}_{q^2}$. Hence every $x\in\mathbb{F}_q\setminus\mathbb{F}_q^2$ has its square roots in $\mathbb{F}_{q^2}$.
This is a general property of finite fields: within a fixed algebraic closure there is exactly one extension of a given degree.
(C) The field $K:=\bigcup\limits_{k\in\mathbb{N}}\mathbb{F}_{q^{2^k}}$ is the smallest square-root-closed field containing $\mathbb{F}_{q}$.
By construction $K$ must be contained in every square-root-closed field containing $\mathbb{F}_{q}$.
(D) Every irreducible polynomial $f\in K[X]$ has odd degree.
Proof: let $x$ be a root of $f$ and choose $k$ such that $K_0:=\mathbb{F}_{q^{2^k}}$ contains all the coefficients of $f$. Then $[K(x):K]=[K_0(x):K_0]$ and $K_0(x).K=K(x)$. Now $K_0(x)/K_0$ is a cyclic extension, because every finite extension of finite fields is cyclic generated by the Frobenius map. Thus if $2$ divides $[K_0(x):K_0]$, there exists an intermediate field $K_0\subset M\subseteq K_0(x)$ such that $[M:K_0]=2$. But then $M\subset K$ and thus $[K(x):K]<[K_0(x):K_0]$.
(E) For every algebraic extension $L$ of $K$ every irreducible polynomial $f\in L[X]$ has odd degree. In particular $L$ is itself square-root-closed.
Proof: let $f\in L[X]$ be irreducible and choose a finite extension $L_0/K$ such that $L_0$ contains the coefficients of $f$. Since $L_0/K$ is separable, its degree by (D) is odd. Let $x$ be a root of $f$. Then the degree of $L_0(x)/K$ is odd too. Consequently the degree of $f$ being equal to the degree of $L_0(x)/L_0$ must be odd.
The following is worth noting: (E) is true (same proof) for every separable extension $L$ of a field $K$ for which every irreducible polynomial has odd degree..
(F) Every square-root-closed algebraic extension of $\mathbb{F}_q$ is an algebraic extension of $K$.
Facts from Valuation Theory
I use the following sources:
[E] O. Endler, Valuation Theory
[K] Franz-Viktor Kuhlmann, http://math.usask.ca/~fvk/Fvkbook.htm
[Z] P. Samuel, O. Zariski, Commutative Algebra II
For convenience I recall some basic definitions and facts:
(1) A valuation $v$ of a field $K$ is a surjective group homomorphism $v:K^\ast\rightarrow\Gamma$, where $\Gamma$ is a totally ordered abelian group (called the value group of $v$), such that $v(x+y)\geq\min (vx,vy)$.
(2) The valuation ring of $v$ is the subring $O:=\{x\in K^\ast : vx\geq 0\}\cup 0$. It is a local ring with maximal ideal $M:=\{x\in K^\ast : vx> 0\}\cup 0$. The field $k(v)=O/M$ is called the residue field of $v$.
(3) A valuation $v$ of $K$ has at least one prolongation $w$ to an extension field $L$ of $K$. The value group $\Delta$ of $w$ contains $\Gamma$ as a subgroup. The residue field $k(w)$ is an extension field of $k(v)$. If $L/K$ is algebraic, then $k(w)/k(v)$ is algebraic and $\Delta /\Gamma$ is a torsion group. [E], Chapter 2, Paragraph 9 or [Z], Chapter 6, Paragraph 4.
(4) A valuation $v$ (or the pair $(K,v)$) is called henselian, if $v$ has exactly one prolongation to any algebraic extension field of $K$.
(5) For every valuation $v$ of a field $K$ there exists an algebraic extension field $K^h$ of $K$ and a prolongation $v^h$ of $v$ to $K^h$ such that $(K^h,v^h)$ is henselian and the value groups and the residue fields of $v$ and $v^h$ coincide. $(K^h,v^h)$ is called the henselisation of $(K,v)$. [E], Chapter 3, Paragraph 17.
(6) Let $(K,v)$ be henselian. Then for every finite extension $L/K$ the following equation holds: $[L:K]=(\Delta :\Gamma )[k(w):k(v)]d(w/v)$, where $w$ is the unique prolongation of $v$ to $L$, $\Delta$ is the value group of $w$, $(\Delta :\Gamma )$ is the index of $\Gamma$ in $\Delta$ and $d(w/v)$ is a power of the characteristic of $k(v)$ (called the defect of $w$). [K], Chapter 11, Lemmas 11.1 and 11.17
(7) Let $K$ be an src-field and $v$ a valuation on $K$, then $\Gamma$ is $2$-divisible (that is $\Gamma =2\Gamma$) and $k(v)$ is an src-field.
If $(K,v)$ is henselian, $\Gamma$ is $2$-divisible and $k(v)$ is an src-field, then $K$ is an src-field.
[K], Chapter 9, Corollary 9.38
Proof of the assertion for fields in class 2 and Remarks
Proof of the Theorem: let $L/K$ be a finite extension and let $w$ be the unique prolongation of $v$ to $L$. By (7) the value group $\Gamma$ of $v$ is $2$-divisible, hence $(\Delta :\Gamma )$ is odd. Since $k(v)$ is an odd field by assumption, the degree $[k(w):k(v)]$ is odd. The defect $d(w/v)$ is odd since $k(v)$ has characteristic $p\neq 2$. Hence by (6) the degree $[L:K]$ is odd.
Proof of the assertion about class 2: by (3) the residue field $k(v)$ is an algebraic extension of a finite field. By (7) $k(v)$ is an src-field. Thus $k(v)$ is in class 1 and therefore an odd field. The theorem now yields the assertion.
Remarks: let $v_0$ be a valuation on a field $K_0$ such that $k(v_0)$ has characteristic $\neq 2$. Given an algebraic extension $k/k(v_0)$ Valuation Theory shows that there exists an algebraic extension $K/K_0$ and a prolongation $v$ of $v_0$ to $K$ such that $k\subseteq k(v)$ and the value group of $v$ is $2$-divisible.
If $K_0$ is not algebraically closed, the field $K$ can be chosen to be rather small compared to the algebraic closure of $K_0$.
Combining these facts with (5) one can prove that class 2 contains more fields than just the algebraic closure of $\mathbb{Q}$: take $K_0$ to be the rationals, $v_0$ a $p$-adic valuation with $p\neq 2$, $k$ an algebraic src-extension field of the finite field $\mathbb{F}_p$. Let $K_1$ be the field obtained by adjoining all elements of the form $p^{1/2^k}$, $k\in\mathbb{N}$ to $K_0$ and let $v_1$ be a prolongation of $v_0$ to $K_1$. Let $K_2$ be an algebraic extension of $K_1$ such that for some prolongation $v_2$ of $v_1$ to $K_2$ the inclusion $k\subseteq k(v_2)$ holds. Finally let $(K,v)$ be the henselisation (5) of $(K_1,v_1)$. Then by construction the value group of $v_1$ and thus of $v$ is $2$-divisible. Moreover $k\subseteq k(v)$, hence $k(v)$ is an src-field. Therefore by (7) $K$ is an src-field.
Another example: let $K$ be the field obtained by adjoining to the Laurent series field $\mathbb{C}((t))$ all elements of the form $t^{1/2^k}$, $k\in\mathbb{N}$. The valuetion $v$ is the unique extension of the natural discrete valuation of $\mathbb{C}((t))$ to $K$.
Lierre's contribution
The proof of the complement indeed shows that the fields in class 2 of my post are odd: let $L/K$ be a finite extension and let $M/K$ be its Galois hull. Then one has the normal series $V\subseteq T\subseteq G$ of the Galois group $G:=\mathrm{Gal}(M/K)$, where $T$ is the inertia group of the prolongation $w$ of $v$ to $M$, and $V$ is its ramification group. Ramification theory shows that $G/T$ is isomorphic to the abelian group $\mathrm{Gal}(k(w)/k(v))$, $T/V$ is abelian and $V$ is a $p$-group. Thus $G$ is solvable. Using Lierre's proof of the complement shows that $[M:K]$ and thus $[L:K]$ are odd.
This way of proving the assertion for the fields of class 2 is much more elegant than mine! ${}$