Why can projection function on $X \times S$ be regarded as a local homeomorphism? I am studying some properties of local homeomorphism 
I am in particular trying to find a local homeomorphism that is not a homeomorphism and the projection function seems to be the perfect candidate since it doesn't "recover". 
I found an example here:

For $X$ any topological space and for $S$ any set regarded as a
  discrete space, the projection
$\pi_X: X\times S \to X$ is a local homeomorphism.

I am trying to see how this function fits the definition: 

Definition: Let $(X, \mathcal{T})$ and $(Y, \mathcal{J})$ be topological spaces. A
  function ${\displaystyle f:X\to Y\,}$  is a local homeomorphism if for
  every point $x \in X$ there exists an open set $U \subseteq X$
  containing $x$ and an open set $V \subseteq Y$ such that the restriction ${\displaystyle f|_{U}:U\to V\,}$  is a  homeomorphism.

So let $V_x$ be an open set on $X$, and $U_x$ some open set in $X \times S$ containing $x$, we wish to show that $\pi_X|_{U_x}: U_x \to V_x $ is a homeomorphism


*

*Show $\pi_X|_{U_x}$ is continuous: Take an open set in $V_x$, then such set has the form $V_x \cap W$, $W$ is open in $X$, $\pi_X|_{U_x}^{-1}(V_x \cap W) = \pi_X|_{U_x}^{-1}(V_x) \cap \pi_X|_{U_x}^{-1}(W) = \pi_X|^{-1}(V_x) \cap  U_x \cap  \pi_X|^{-1}(W)$, right handside is open in the product topology on $X \times S$ 

*Show $\pi_X|_{U_x}$ is open: Take an open set in $U_x$, then such set has the form $U_x \cap M$, $M$ is open in $X \times S$, $\pi_X|_{U_x}(U_x \cap M) = \pi_X|_{U_x}(U_x) \cap \pi_X|_{U_x}(M) = \pi_X( U_x) \cap\pi_X(M) \cap \pi_X( U_x)$, right handside is open since $\pi_X$ is open map$ 
[^ actually I noticed that we cannot do $f(A \cap B) = f(A) \cap f(B)$ since $\pi$ is not a bijection...]
Further, this map is not a homeomorphism since $\pi_X$ does not biject.

Is the above good enough? Is there any easier way to show this result?

Note: another example that seems more accessible is $f: \mathbb{C} \to \mathbb{C}$, $f(z) = e^z$, but I don't know about topologies on complex spaces
 A: You have given an argument that $\pi_X|_{U_x}$ is continuous and open, but you are missing the most important part: a homeomorphism needs to be a bijection!  In fact, $\pi_X|_{U_x}$ won't be a bijection if you choose $U_x$ and $V_x$ arbitrarily (for instance, you might choose $U_x$ to be all of $X\times S$ and $V_x$ to be all of $X$, and then $\pi_x$ is not a bijection unless $S$ has exactly one point.)
So you need to use the fact that the definition of "local homeomorphism" lets you choose $U$ and $V$.  Following the definition, you should not start with $V_x$ and $U_x$, but only with a point $(x,s)\in X\times S$.  You then get to choose an open set $U$ containing $(x,s)$ and an open set $V$ such that $\pi_X$ will restrict to a homeomorphism $U\to V$.  Let's choose $V=X$ and $U=X\times \{s\}$.  Note that $U$ is open because $S$ has the discrete topology so $\{s\}$ is open in $S$.  Then it is easy to show that $\pi_X$ restricts to a bijection $U\to V$, and your argument shows this bijection is a homeomorphism.
In fact, if you just want any example at all, there is a super-simple one: just consider the map $\emptyset\to Y$ for any nonempty space $Y$.  This is trivially a local homeomorphism, since there are no points in the domain to choose when checking the definition!
A: The sets $X_s = X \times \{s\} \subset X \times S$ are all open since $S$ is discrete, and $\pi_X|_{X_s} \colon X_s \to X$ is a homeomorphism (the inverse is $x \mapsto (x,s)$). Since $\{X_s\}_{s\in S}$ form an open cover of $X\times S$, we have shown $\pi_X$ is a local homeomorphism. 
