Tychonoff's theorem:$\phantom{---}$ If $A$ is a non-empty index set and $X_{\alpha}$ is a non-empty compact topological space for every $\alpha\in A$, then $X\equiv\times_{\alpha\in A} X_{\alpha}$ is compact in the product topology.
This theorem is well-known to be equivalent to the axiom of choice.
Converse of Tychonoff's theorem:$\phantom{---}$ If $X$ is not empty and compact in the product topology, then $X_{\alpha}$ is compact for every $\alpha\in A$.
Proof:$\phantom{---}$ Pick any $\alpha\in A$ and $x_{\alpha}\in X_{\alpha}$. For any other $\beta\in A\setminus\{\alpha\}$, pick an arbitrary $x_{\beta}\in X_{\beta}$. Construct $\mathbf x\in X$ such that $\pi_{\alpha}(\mathbf x)=x_{\alpha}$ and $\pi_{\beta}(\mathbf x)=x_{\beta}$ for any $\beta\in A\setminus\{\alpha\}$, where the $\pi_{\cdot}(\cdot)$ denote the coordinate maps. It follows that $x_{\alpha}\in \pi_{\alpha}(X)$, and, in turn, that $X_{\alpha}=\pi_{\alpha}(X)$. Since the coordinate maps are continuous by the very construction of the product topology and $X$ is compact, $X_{\alpha}$ is compact.$\phantom{---}\blacksquare$
Notice that this easy proof makes use of the axiom of choice, which guarantees the existence of $\mathbf x$. What I am wondering about is whether the converse of Tychonoff's theorem can be proved in ZF without AC (or weaker variants of it)?
Thank you.
If $X$ is non-empty, then there is no dependence on the axiom of choice. To see this, note that $X_\alpha$ is a continuous map of $X$ with the projection map $\pi=\pi_\alpha(x)=x_\alpha$.
This follows from the fact that if a product $X=\prod_{i\in I}X_i$ is non-empty, then for each $x\in X_i$ there is a function $f\in X$ with $f(i)=x$. Simply pick one element $g\in X$, which exists since $X$ is non-empty, then for each $x\in X_i$ define $f_x=(g\setminus\{\langle i,g(i)\rangle\})\cup\{\langle i,x\rangle\}$.
This $\pi$ is continuous since given $U\subseteq X_\alpha$ which is open, the preimage of $U$ is the product of $U_i$'s where $U_\alpha=U$ and $U_i=X_i$ for $i\neq\alpha$.
And one can easily show that the continuous image of a compact space is compact without using the axiom of choice.
If you allow $X$ to be empty, this is of course equivalent to the axiom of choice. Pick any infinite family of infinite sets whose product is empty, and consider them with discrete topologies.