There is a very natural proof that the product of path-connected spaces is path-connected:
Let $X=\prod_{i\in I}X_i$ be a product of path-connected spaces $X_i$. Given $(x_i)_{i\in I},(y_i)_{i\in I}\in X$, by assumption for each $i$ there is a path $\gamma_i$ from $x_i$ to $y_i$. Then the function $\gamma(t)=(\gamma_i(t))_{i\in I}$ is a path from $(x_i)_{i\in I}$ to $(y_i)_{i\in I}$.
This proof uses the (full) axiom of choice in the selection of the $\gamma_i$'s. Now what I would like to know is whether, as with Tychonoff's theorem, you can recover AC from this statement.
Yes! Assume that any product of path-connected spaces is path-connected. Let $(X_i)_{i\in I}$ be an arbitrary collection of nonempty sets. We want to demonstrate a choice function on $(X_i)$.
Equip $X_i$ with the discrete topology. Let $SX_i$ be the suspension of $X_i$: explicitly, this is the quotient of $X_i\times[0,1]$ by collapsing the edges $X_i\times\{0\}$ and $X_i\times \{1\}$ to points we will call $0_i$ and $1_i$. Since $X_i$ is nonempty, $SX_i$ is path-connected. Now by assumption, $\prod_iSX_i$ is path-connected. Let $\gamma:[0,1]\to\prod_iSX_i$ be a path from $\prod_i0_i$ to $\prod_i1_i$. I claim that $\gamma$ defines a choice function, as follows.
Let $\gamma_i:[0,1]\to SX_i$ be the induced path on $SX_i$. Let $t_i\in(0,1)$ be the least time such that the second coordinate of $\gamma_i(t_i)$ equals $1/2$. The first coordinate of $\gamma_i(t_i)$ is an element $x_i\in X_i$. We're done: $i\mapsto x_i$ is our choice function.