Prove that if $A$ is an infinite set then $A \times 2$ is equipotent to $A$

Consider the collection $S$ of pairs $(X,f)$ where $X\subseteq A$ is infinite and $f:X\times 2\hookrightarrow X$. Then $S$ is nonempty, since $A$ being infinite contains a copy of $\Bbb N$, and it is known $\Bbb N\times 2\simeq \Bbb N$. It should be evident that under the ordering of extension, this set always has $(X,f)\in S$ if $X=\bigcup X_i$ and $f=\bigcup f_i$ with each $(X_i,f_i)\in S, i\in I$ and $I$ a total order. It follows by Zorn's lemma that there is a maximal element $(A',g)$ in $S$. If we show that $A'$ has the same cardinality as $A$, we're done. Now it is clear $\# A'\leqslant \# A$ since $A'$ is a subset of $A$. So assume $\#A'<\#A$. Then $A\setminus A'$ must be infinite, since else we would have by $A=A'\sqcup (A\setminus A')$ an equality (recall that if $\mathfrak a$ is infinite and $\mathfrak b$ finite, $\mathfrak a+\mathfrak b=\mathfrak a$) , so there is some $A''\subset A$ disjoint from $A'$ with cardinality $\aleph_0$. But then by patching things together, we would be able to extend the injection $A'\times 2\to A'$ to a larger $(A'\sqcup A'')\times 2\to (A'\sqcup A'')$.

Here's a kind of brute-force way to do it.

Lemma. Assume that $A$ is infinite. Then there is a set $B$ such that $A$ is equipotent to $B\times\mathbb N$.

Assuming this lemma, we only need to prove your property when $A$ is in fact $B\times \mathbb N$. But then, obviously, this is a bijection: $$ f:B\times\mathbb N\times \{0,1\} \to B\times\mathbb N : (b,n,m)\mapsto(b,2n+m) $$

Proof of lemma. Let a "cool" function mean an injective function $g:B\times \mathbb N\to A$ for some $B\subseteq A$, where $g(b,0)=b$ for all $b\in B$. Order the cool functions by set inclusion; Zorn's lemma then applies, and we get a cool function $g$ that cannot be extended to a larger cool function.

Now if $A\setminus \operatorname{Rng} g$ is infinite, then it has a countably infinite subset (I assume this is known, as a consequence of AC or Zorn), and then it is easy to see that $g$ cannot have been maximal. (Here it is useful that $g(b,0)=b$ such that $B$ is disjoint from $A\setminus \operatorname{Rng} g$). So $A\setminus \operatorname{Rng} g)$ is finite; call its size $M$.

Let therefore $h:\{0,1,\ldots,M-1\}\to A\setminus \operatorname{Rng} g$ be a bijection. Furthermore $B$ must be nonempty (because $A$ is infinite), choose a $b_0\in B$. Now $$ (b,n) \mapsto \begin{cases} g(b,n) & \text{if }b\ne b_0 \\ h(n) & \text{if }b=b_0 \land n<M \\ g(b,n-M) & \text{if }b=b_0 \land n\ge M \end{cases}$$ is a bijection $B\times\mathbb N\to A$.