Continuity of projection map
Suppose $W \subset X$ is open and $V \subset Y$ is not (e.g., is closed). Projection $\pi:X \times Y \to X$ sends the non-open set $W \times V$ to the open set $W$. Context suggests there are two points of confusion:
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As Qiyu Wen notes in the comments, the preimage $\pi^{*}(W) = \pi^{*}(\pi(W \times V))$ is $W \times Y$, which is open in $X \times Y$.
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What if we restrict the projection to $X \times V$? Now the preimage of $W$ is $W \times V$, but this is still (relatively) open in $X \times V$.
Either way, the preimage of an open set is open.
You have to know that if $f: X \to Y$ is any function and $A \subseteq Y$ then $f^{-1}(A)$ is defined as $$\{x \in X\mid f(x) \in A\}$$ i.e. the set of all points $x$ in the domain of $f$ whose $f$-image lies in $A$. So that's a single condition to determine whether a domain point is in $f^{-1}(A)$ or not.
If $U \subseteq X$, then as $\pi(x,y)=x$ and $\pi: X \times Y \to X$ we apply the above definition and look for all pairs $(x,y) \in X \times Y$ so that $\pi(x,y)=x$ lies in $U$. This happens iff $x \in U$. So by definition $\pi^{-1}(U)= U \times Y$ which is exactly the set of all pairs of points with first coordinate in $U$. There is no choosing a $V$, we must include all pairs that satisfy the condition not just those where impose our own arbitary extra conditions... Computing $f^{-1}(A)$ from $A$ is a deterministic process!