I heard the statement from the title of this question. $\newcommand{\of}[1]{\left(#1\right)}$ $\newcommand{\df}{\mathrel{\mathop:}=}$
So if I am not totally confused, this formalizes to:
Let $p\neq 2$ and $u\in\mathbb{Z}_p^*$ with canonical form $\sum_{i=0}^\infty u_ip^i$. Then \begin{align*} \exists\ x\in\mathbb{Z}_p:\ x^2=u \quad \Leftrightarrow \quad u_0 \ \text{is a quadratic residue modulo} p. \end{align*}
I think that the direction "$ \Rightarrow $" is straightforward:
Let $x\in\mathbb{Z}_p$ with canonical form $\sum_{i=0}^\infty x_ip^i\in\mathbb{Z}_p$ such that $x^2 =u$. By the Cauchy-Product-Fomrula, we have that \begin{align*} x^2=\of{\sum_{i=0}^\infty x_ip^i}^2=\sum_{i=0}^\infty \sum_{j=0}^i x_{i-j} x_jp^i\stackrel{!}{=}\sum_{i=0}^\infty u_ip^i. \end{align*} Comparing the coefficient of $p^0$ yields $x_0^2=u_0$ which means that $u_0$ is a quadratic residue modulo $p$.
Now it is left to prove the direction "$ \Leftarrow $" and that's where I get stuck. Here is what I have:
Define $f(X)\df X^2-u_0$. Since $u_0$ is a quadratic residue modulo $p$, we have that \begin{align*} \exists\ 0
$f(y)\equiv 0\pmod{p}$ and since $\gcd\of{y,p}=1$ and $\gcd\of{p,2}=1$, we get that $f'(y)=2y\not\equiv 0\pmod{p}$. So the conditions for Hensel's lemma are fullfilled and we get a unique $a\in\mathbb{Z}_p$ with $a\equiv y\pmod{p}$ and $f(a)=0$.
I am quite sure that I need to use the condition that $u$ is a unit i.e. that $\left|| u \right||_p=1$ or that $u^{-1}\in\mathbb{Z}_p$ but I don't know how. Or maybe I got the statement wrong and am missing any premises?