Show that X $\times$ Y is compact if X and Y are compact. Show that X $\times$ Y is compact if  X and Y are compact. 
I know there are different solutions in this site to this question, but I want to use this exact statement: 
Suppose X and Y are two topological spaces with Y compact, and $x_0$ a point in X. If M is an open set in X $\times$ Y, such that {$x_0$} $\times$ Y $\subseteq$ M, then there exists an open set W $\subseteq$ X containing $x_0$ such that W $\times$ Y $\subseteq$ M
(Using this statement, I have to show that for all x $\in$ X, we can create a tube (covered by finite amount of open sets) around {x} $\times$ Y)
So I start with an open cover for X$\times$Y, say {U$_\alpha$}$_{\alpha\in J}$ = {A$_{\alpha}$$\times$B$_{\alpha}$} where each A$_{\alpha}$ is open in X and each B$_{\alpha}$ is open in Y. What do I do next?  
I know that for any open cover for X, say {A$_\alpha$}$_{\alpha\in J}$, there exists a finite subcover of X. Similarly for Y. 
 A: Sets of the form $U \times V$ where $U$ is open in $X$ and $V$ is open in $Y$ form a basis for the product topology.
Hence $M = \cup_\alpha (U_\alpha \times V_\alpha)$, for some collection of open sets $\{ U_\alpha \times V_\alpha \}_\alpha$.
Let ${\cal A} = \{\alpha | (\{x_0\} \times Y) \cap (U_\alpha \times V_\alpha) \neq \emptyset  \}$. Note that $Y \subset \cup_{\alpha \in {\cal A}} V_\alpha$, and $x_0 \in U_\alpha$ for all $\alpha \in {\cal A}$.
Since $Y$ is compact, we have a finite $I \subset {\cal A}$ such that $Y \subset \cup_{\alpha \in I} V_\alpha$ (hence in fact, $Y=\cup_{\alpha \in I} V_\alpha$), and since $I$ is finite, the set $W = \cap_{\alpha \in I} U_\alpha$ is open, and $x_0 \in W$. Furthermore, $W \times V_\alpha \subset U_\alpha \times V_\alpha$ for all $ \alpha \in I$, and so $\cup_{\alpha \in I} (W \times V_\alpha) = W \times (\cup_{\alpha \in I} V_\alpha) = W \times Y \subset \cup_\alpha (U_\alpha \times V_\alpha) = M$.
To show $X \times Y$ is compact:
Let $\Pi_Y: X \times Y \to Y$ be given by $\Pi_Y((x,y)) = y$. Then if $Z$ is open in $X \times Y$, $\Pi_Y(Z)$ is open in $Y$. To see this, suppose $\Pi_Y((x,y)) = y$. Since $Z$ is open, and products of open sets form a basis for the product topology, we have some $U$ open in $X$ and $V$ open in $Y$ such that $(x,y) \in U \times V \subset Z$. Since $\Pi_Y(U \times V) = V$ and $y \in V$, we see that $\Pi_Y(Z)$ is open in $Y$.
It should be clear that the corresponding $\Pi_X((x,y)) = x$ is also an open map.
Suppose $\{ Z_\alpha \}$ is an open cover of $X \times Y$. Pick some $x_0 \in X$. Let ${\cal A_{x_0}} = \{\alpha | (x_0,y) \in Z_\alpha \text{ for some } y \in Y \} $. Since $\{ Z_\alpha \}$ covers $X \times Y$, we see that$ \{ \Pi_Y(Z_\alpha) \}_{\alpha \in {\cal A_{x_0}}}$ is an open cover of $Y$, and so there is a finite subcover $\{ \Pi_Y(Z_\alpha) \}_{\alpha \in {I_{x_0}}}$, where $I_{x_0}$ is finite. The above result shows that there is some open $W_{x_0}$ such that $W_{x_0} \times Y$ is covered by $\{ Z_\alpha \}_{\alpha \in {I_{x_0}}}$.
The sets $\{W_{x_0} \}_{x_0 \in X}$ form an open cover of $X$, hence there is a finite subcover $\{ W_{x_0} \}_{x_0 \in F}$, where $F \subset X$ is finite.
Now take the collection $\{ Z_\alpha \}_{x_0 \in F, \alpha \in {I_{x_0}} }  $ and note that it covers $X \times Y$ and is finite. Hence $X \times Y$ is compact.
A: Not sure if this is what you are looking for : You can assume that $U_{\alpha} = V_{\alpha}\times W_{\alpha}$ is a basic open set for all $\alpha \in J$. For each $x\in X$, $\{x\}\times Y$ can be covered by finitely many such $V_{\alpha}\times W_{\alpha}$. Take their intersection to see that there is a neighbourhood $V_x$ of $x$ such that $V_x\times Y$ is covered by finitely many such $V_{\alpha}\times W_{\alpha}$.
Now $\{U_x : x\in X\}$ covers $X$. Take a finite subcover, and you should be done.
