It's before we prove that 'Product of two compact sets is compact'. Let $S$ be an open cover of $X \times \{\bullet\}$ where $X$ is compact. Then $\pi_1(S)$ is an open cover of $X$ so there is a finite subcover then pick one $A_n$ from $S$ corresponding to each $\pi_1(A_n)$ but I think it may not cover $X \times \{\bullet\}$. How to complete the proof?
Edit There are topological spaces $X$(which is compact), $Y$ and the product topology on $X \times Y$ is given by the subbase $U \times V$ where $U$ is open in $X$ and $V$ is open in $Y$. Pick an element $\bullet \in Y$. Let $S$ be an open cover of $X \times \{\bullet\}$. Then $\pi_1(S)$ is an open cover of $X$ so there is a finite subcover then pick one $A_n$ from $S$ corresponding to each $\pi_1(A_n)$ but I think it may not cover $X \times \{\bullet\}$. For example let $Y$ be a $T_1$ space and pick another element $\bullet\bullet \in Y$. There exists an open set $W$ containing $\bullet\bullet$ but not $\bullet$. Lets construct an open cover of $X \times \{\bullet\}$ $$S \cup \{B \times W \mid B \in \pi_1(S)\}$$ Now we can apply this open cover to upper proof, and when we are picking $A_n$ from $S$ corresponding to each $\pi_1(A_n)$, we may pick all the sets from $\{B \times W \mid B \in \pi_1(S)\}$ so in fact it doesn't cover $X \times \{\bullet\}$. Am I misunderstanding something?
It will cover $X \times \{∙\}$.
You have a finite collection $\pi_1(A_1),\dots,\pi_1(A_n)$ of open subsets of $X$ that cover $X$. Choose any $(x,∙) \in A_i$. Then $x \in X$, so there exists $A_i$ such that $x \in \pi_1(A_i)$. This means there exists $(y,z) \in A_i\subseteq X \times \{∙\}$ such that $\pi_1(y,z)=x$. But the only $z \in \{∙\}$ is $∙$ itself, and $\pi_1(y,z)=y$ by definition. So we have $(x,∙) \in A_i$.
However, you should also justify why $\pi_1(A_i)$ is open. In general the image of an open set is not necessarily open, but it will hold for projections.