If $X$ is not compact, then $X\times Y$ is not compact Exercise If $X,Y $are two topological spaces and $X$ is not compact then, $Y$ is any space then $X \times Y$ is not compact.
My attempt: So we know that topological space is compact if every open cover has a finite subcover, so take an open cover of $X \times Y$. Say $U_\alpha$. We know that each open set in $X\times Y$ is of the form $U_i = V \times V'$, where V is open in X and V' is open in Y. Hence $U_\alpha = V_\alpha \times V'_\alpha$ ,where $V_\alpha $ and $V'_\alpha$ are open covers of X and Y respectively. But since X and Y are not compact we know that there does not exist finite subcover of neither $V_\alpha $ nor $V'_\alpha$. Hence we can not find finite subcover of $V_\alpha \times V'_\alpha$ and hence for $U_\alpha$.
Is it enough or it is not sufficient?
 A: "We know that each open in $X\times Y$ is of the form..." This isn't true. The open unit disc $\{x: |x|<1\}$ is open in $\Bbb R\times\Bbb R$, for instance.
Hint: If you want to prove compactness, you need to take an arbitrary cover. If you want to disprove compactness, you can choose a cover to your liking. Use the non-compactness of $X$ to construct a cover of $X\times Y$ such that...
A: It is not true that every open set in $X\times Y$ is of the form $V\times V'$; for the easiest example, take the unit disk in $\mathbb R^2$. 
Also, you won't get anywhere by starting with some open cover. You want to show that something is not compact, you need to exhibit a single open cover that does not admit  a finite subcover. So, using that $X$ is not compact, you know that there exists an open cover with no finite subcover; now use this information to construct an open cover of $X\times Y$ that does not admit an open subcover. 
A: If $X \times Y$ were compact, then $X = \pi_X[X \times Y]$ is also compact, as
the continuous image of a compact space under the continuous projection map $\pi_X: X \times Y \to X$ defined by $\pi_X(x,y) = x$.
Using covers: let $U_\alpha, \alpha \in A$ be an open cover of $X$ without a finite subcover.
Then $\{U_\alpha \times Y: \alpha \in A\}$ is an open cover of $X \times Y$ without a finite subcover (if $Y\neq \emptyset$). So $X \times Y$ is not compact.
