Projections are open maps. Why might I be wrong? I got this problem from Munkres, my idea is similar, but comparing to the actual solution, I missed at least 4 steps.

Prove that the projection maps $\pi_1 : X \times Y \to X$ and $\pi_2 : X \times Y \to Y$ are both open maps.

Here is my solution

Let $U_1 \times U_2 \subset X \times Y$ be open in $X \times Y$, meaning $U_1$ and $U_2$ are open in $X$ and $Y$ respectively that is $U_1 \in \tau_X$ and $U_2 \in \tau_Y$. Now by the definition of $\pi_1$, we have $$\pi_1(U_1 \times U_2) = U_1 \in \tau_X$$ so the image is also open. The proof for $\pi_2$ is similar.

Now the solution in page 4 of the pdf includes all these union and basis things. The only major thing I noticed is that I didn't consider either $U_1$ or $U_2$ to be empty, but I am not sure how meaningful it is to project on empty sets because the projection is still empty and that is clearly open.
 A: I think the main issue with your solution is that you consider only those open sets in the product topology that are of the form $U_1\times U_2$ for some $U_1$ and $U_2$ open in $X$ and $Y$, respectively. Sure, sets of this form are open in the product topology, but there are many more open sets that cannot be expressed as simple products.
In particular, $V\subseteq X\times Y$ is open in the product topology if and only if there exist collections of open sets $\{U_1^{\alpha}\}_{\alpha\in A}$ in $X$ and $\{U_2^{\alpha}\}_{\alpha\in A}$ in $Y$, where $A$ is a non-empty index set, such that $$V=\bigcup_{\alpha\in A}U_1^{\alpha}\times U_2^{\alpha}.$$
As for the issue with empty sets, if either $U_1$ or $U_2$ is empty, then $U_1\times U_2$ is empty and the image of the empty set is vacuously empty.
A: As I noted in the comments of triple_sec's answer, what you did is in fact sufficient to prove that projections are open mappings. The reason your proof is enough is that a basis for the product topology on $X\times Y$ is given by sets of the form $U_1\times U_2$, with $U_1\subseteq X$ open and $U_2\subseteq Y$ open. In general, it is true that $f(A\cup B) = f(A)\cup f(B)$, so it is sufficient to check openness of a map $f : X\to Y$ on a basis for $X$ (i.e., check that any basis open has open image in $Y$ under $f$), the same way that it is sufficient to check continuity of $f$ on a basis of $Y$ (check that any basis open in $Y$ has open preimage in $X$). Of course, if you are proving statements like "a projection map is open," you will probably be expected to say why it's OK that you only check openness on a basis.
However, one does need to be careful, as a basis $\beta$ for a topology $\tau$ on $X$ is not the whole story: a general open set in $\tau$ is some union of basis opens, but not necessarily an element of $\beta$ itself! Although there are a number of properties that can be checked on a basis, there are also properties that cannot be checked on a basis, so one must realize the difference between $\beta$ and $\tau$ when examining a space $X$.
