Let $X$ and $Y$ be two locally connected spaces. I need to show that $Z = X \times Y$ is locally connected. Here is my attempt:
Proof. Let $z=(x, y)\in Z$ and $N$ be any neighborhood of $z$. I need to show that $\exists U\subset N, U$ is connected. Now, since $Z = X\times Y, N_x\times N_y\subset N$, where $N_x$ is a neighborhood of $x$ and $N_y$ is a neighborhood of $Y$. Since $X$ is locally connected, $\exists U_x\subset N_x, U_x$ is connected. Similarly, $\exists U_y\subset N_y, U_y$ is connected. Now let $U=U_x\times U_y$.
I tried prove that $U$ must be connected by contradiction but to no avail. Maybe the correct approach would be to show that connectivity of $U$ follows from the fact that it is a product of connected sets but I am not sure how to make such a leap. Any help would be appreciated.
If U were not connected, there would be a set $F \subset U $ that was closed and open; then, the projection of that set down to either X or Y would also be clopen (because the projection is continuous); but X and Y are locally connected, which means that $\pi_X(F)=U_X, \pi_Y(F)=U_Y$ (or the empty set of course, but that alternative is analogous). Then, projecting back, you get that $F=U$; similarly for the case where either the projections were empty, $F=\emptyset$.
That is, you proved that the only clopen subset of $U$ is either $U$ itself or the empty set, ie $U$ is connected.