I'm guessing that what you wanted to ask is the following:
Let $K$ be a field, and let $L$ be an extension of $K$ (that is, $K\subseteq L$). Suppose futher that $L$ is algebraically closed (all non-constant polynomials in $L[x]$ have a root in $L$). Is the set $\{a\in L\mid a\text{ is algebraic over }K\}$algebraically closed?
The answer is "yes."
Let $F = \{ a\in L\mid a\text{ is algebraic over }K\}$, and let $f(x)\in F[x]$ be a nonconstant polynomial. We may assume $f(x)$ is monic, $f(x) = x^n + a_{n-1}x^{n-1} + \cdots + a_0,\quad a_i\in F,\quad n\gt 0.$ Since $a_i\in F$, then $a_i$ is algebraic over $K$ for all $i$, so the field $E=K(a_0,\ldots,a_{n-1})$ is a finite extension of $K$.
Now $f(x)\in F[x]\subseteq L[x]$, so we know that $L$ contains a root $r$ of $f(x)$, since $L$ is algebraically closed. Therefore, $[E(r):E] \leq n$ is finite, so $[E(r):K] = [E(r):E][E:K]$ is also finite. Therefore, $[K(r):K] \leq [E(r):K]$ is finite, so $r$ is algebraic over $K$. Since $r$ is algebraic over $K$, then $r\in F$. Thus, $f(x)$ has a root in $F$.
This shows that every nonconstant polynomial in $F(x)$ has a root in $F$, so $F$ is algebraically closed.
Apparently not. Rather, you want:
Let $K$ be a field, and let $L$ be an extension of $K$ (that is, $K\subseteq L$). Suppose further that all non-constant polynomials in $K[x]$ have a root in $L$. Is the set $\{a\in L\mid a\text{ is algebraic over }K\}$algebraically closed?
This is equivalent to asking whether an algebraic extension of $K$ in which every polynomial in $K$ has a root is algebraically closed (since the subfield of $L$ of all elements that are algebraic over $K$ contains all the roots of polynomials in $K[x]$).
Again, the answer is yes.
Take $L$, $F$, and $f$ as above. Since $K(a_0,\ldots,a_{n-1})$ is algebraic over $K$, and $f(x)$ is a polynomial in $K(a_0,\ldots,a_{n-1})$, then the roots of $f(x)$ are algebraic over $K$. Thus, if $r$ is a root of $f(x)$ in some algebraic closure of $F$, then $r$ has an irreducible polynomial $g(x)\in K[x]$. Let $h(x)$ be the irreducible polynomial of $r$ in $F[x]$. In particular, $h(x)$ divides both $f(x)$ and $g(x)$ in $F[x]$. Since $g(x)$ has all its roots in $L$, and all its roots lie in $F$, then $g(x)$ has all its roots in $F$. Moreover, $h(x)|g(x)$ in $F$, so every root of $h(x)$ is a root of $g(x)$, hence every root of $h(x)$ lies in $F$. But $h(x)$ divides $f(x)$, so $f(x)$ has a root in $F$. Therefore, $F$ is algebraically closed.
Added. Hmph. Wait a second here. Every polynomial in $K[x]$ has at least one root in $L$, but we're not asserting they split. "Every polynomial in $L$ has at least one root in $L$" is equivalent to "every polynomial in $L$ splits", but this is not what we have here. We would need to show that
If $L$ is an algebraic extension of $K$ such that every every nonconstant polynomial in $K[x]$ has a at least one root in $L$, then every nonconstant polynomial in $K[x]$ splits over $L$.
On reflection, I'm not so sure this is clear.
Let $K=\mathbb{Q}$, and let $f(x)=x^3-2$. Consider the collection $\mathfrak{F}$ of all subfield of $\mathbb{C}$ that contain $\sqrt[3]{2}$, but do not contain the complex roots of $f(x)$. The collection is nonempty (it contains $\mathbb{Q}(\sqrt[3]{2})$). Partially order $\mathfrak{F}$ by inclusion. If we have a chain $\mathcal{C}$ of elements of $\mathfrak{F}$, $\mathcal{C}=\{F_i\}_{i\in I}$, then $\cup\mathcal{C}$ is a field, contained in $\mathbb{C}$, and does not contain the complex roots of $f(x)$. By Zorn's Lemma, $\mathfrak{F}$ contains maximal elements. Let $E$ be a maximal element of $\mathfrak{F}$.
Does every irreducible nonconstant polynomial in $\mathbb{Q}[x]$ have at least one root in $E$?
On further reflection, the result is definitely true if $K$ is perfect (in particular, in characteristic zero, which takes care of my stricken attempt at an example above). I think it is true in general (by first closing under adjunction of $p$th roots and then using the result for perfect fields to get the conclusion), but I haven't quite worked out all the details yet.
Theorem. Let $L$ be an extension of $K$, and assume that every separable nonconstant polynomial in $K[x]$ has at least one root in $L$. Then every nonconstant separable polynomial in $K[x]$ splits over $L$.
Proof. Let $f(x)$ be a separable nonconstant polynomial in $K[x]$, and let $r\in L$ be a root of $f(x)$. Let $\overline{K}$ be an algebraic closure of $K$. Then there is an embedding of the subfield $K(r)$ of $L$ into $\overline{K}$, so we may assume $K(r)\subseteq \overline{K}$. Let $F$ be the splitting field of $f(x)$ over $K$. Since $F$ is separable and of finite degree over $K$, by the Primitive Element Theorem, there exists $a\in\overline{K}$ such that $F=K(a)$. Let $g(x)$ be the irreducible polynomial of $a$ over $K$. Then we know that $g(x)$ has a root $\alpha\in L$, and that the subfield $K(\alpha)$ of $L$ embeds into $\overline{K}$. Since $K(a)$ is normal, we must have that the image of $K(\alpha)$ in $\overline{K}$ is $K(a)$; that is, $K(\alpha)\cong K(a)$. But $f(x)$ splits in $K(a)$, hence $f(x)$ splits in $K(\alpha)\subseteq L$ Thus, $f(x)$ splits in $L$, as claimed. QED
Corollary. Let $K$ be a field, and let $L$ be an extension of $K$ such that every nonconstant polynomial in $K[x]$ has at least one root in $L$. If $K$ is perfect, then the subfield $E=\{a\in L\mid a\text{ is algebraic over }K\}$ is algebraically closed.
Proof. If $f(x)$ is any nonconstant polynomial in $K[x]$, then $f(x)$ is separable (since $K$ is perfect). Since $L$ satisfies the hypothesis of the theorem, then $f(x)$ splits over $L$, hence over $E$. Thus, $E$ is a splitting field of all nonconstant polynomials in $K[x]$, and algebraic over $K$, hence $E$ is algebraically closed. QED
Final addition: And Keith Conrad has been kind enough to point us in the comments to his notes on the subject, which shows that, indeed, first adjoining all the
$p^k$th roots to all elements of
$K$ will work to give the result in general. So the result
does hold: if every polynomial in
$K[x]$ has at least one root in
$L$, then
$L$ contains an algebraic closure for
$K$.