It is known that the only elementary abelian $2$-groups (finite and nonfinite) in $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ are in fact finite and cyclic – that is to say, they are of order $2$.
(Here, $\overline{\mathbb{Q}}$ is the algebraic closure of $\mathbb{Q}$ inside $\mathbb{C}$.)
Question: Does one need the Axiom of Choice to acquire this property ?
No, you shouldn't need any choice for this, and it should still be true if you replace $\overline{\mathbb{Q}}$ with any other algebraic closure of $\mathbb{Q}$. Let $K$ be a field (which in our application will be $\overline{\mathbb{Q}}$) and let $G$ be a finite group of automorphisms of $K$. Then $K/K^G$ is always a degree $\#(G)$ extension. (See, e.g., Milne - Fields and Galois theory Corollary 2.14 and Theorem 3.4.)
The Artin–Schreier theorem says that, if $K/F$ is a finite degree extension and $K$ is algebraically closed, then $[K:F] \leq 2$.
Combining the two statements, if $G$ is a finite subgroup of $\operatorname{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$, then $\#(G) \leq 2$.
There is no use of choice in the proof of either of the big tools that I'm using.
Let me spell out a completely explicit elementary proof that visibly makes no use of choice.
Lemma 1. Let $\sigma$ be an automorphism of order $2$ of a field $K$ of characteristic $\ne2$, and let $F$ be the fixed field of $\sigma$. Then $K=F(\sqrt a)$ for some $a\in F$.
Proof: Since $\sigma$ is not the indentity, $K\ne F$, thus fix $\alpha\in K\smallsetminus F$. We have $\alpha+\sigma(\alpha),\alpha\sigma(\alpha)\in F$, hence $\alpha$ is quadratic over $F$; using the quadratic formula, we may assume $\alpha=\sqrt a$ for some $a\in F$.
If $\beta\in K$ is arbitrary, the same argument shows that $\beta\in F(\sqrt b)\subseteq K$ for some $b\in F$. Either $\sqrt b\in F$, or $\sigma(\sqrt b)=-\sqrt b$ and $\sigma(\alpha)=-\alpha$, hence $\sqrt b/\alpha\in F$. In both cases, $\beta\in F(\alpha)$. QED
Lemma 2. If, furthermore, $K$ is quadratically closed, then $K=F(\sqrt{-1})$, and for every $w\in F$, $w$ or $-w$ has a square root in $F$.
Proof: By assumption, $\alpha=\sqrt a$ has a square root in $K$, i.e., $\alpha=(u+\alpha v)^2=(u^2+av^2)+2uv\alpha$ for some $u,v\in F$. Then $u^2+av^2=0$, and since $v\ne0$, we see that $\sqrt{-a}\in F$, hence $i=\sqrt{-1}\notin F$, hence $K=F(i)$ as $[K:F]=2$.
Any $w\in F$ has a square root in $K$, thus $w=(u+iv)^2=(u^2-v^2)+2uv i$ for some $u,v\in F$. Then $2uv=0$, thus $u$ or $v$ is $0$, thus $w=u^2$ or $-w=v^2$. QED
Theorem. If $K$ is a quadratically closed field of characteristic $\ne2$, and $\sigma$ and $\tau$ are commuting automorphisms of $K$ of order $2$, then $\sigma=\tau$. Consequently, $\mathrm{Aut}(K)$ has no elementary abelian $2$-subgroup of order $>2$.
Proof: Let $F$ be the fixed field of $\sigma$. Then $K=F(i)$ by Lemma 2. Since $\tau$ commutes with $\sigma$, it restricts to an automorphism of $F$. If $\tau$ is the identity on $F$, then $\sigma=\tau$ is the unique automorphism that fixes $F$ and negates $i$. Otherwise, let $H$ be the fixed field of $\tau$ within $F$; we have $F=H(\alpha)$, $\alpha=\sqrt a$ for some $a\in H$ by Lemma 1. By Lemma 2, we may assume $\alpha$ has a square root in $F$, thus $\alpha=(u+\alpha v)^2=(u^2+av^2)+2uv\alpha$ for some $u,v\in H$. Then as above, $u^2+av^2=0$ implies that $\sqrt{-a}\in H\subseteq F$. But then $\sqrt{-1}\in F$ and $K=F$, a contradiction. QED
