I think I have solved a problem in Topology by Munkres, but there is a small detail that is bugging me. The problem is stated in this question's title. I will write down the proof and will highlight what is troubling me.
We prove by contradiction: Assume $X$ is not Hausdorff. Then there exist points $x,y$ where $x$ is different from $y$ such that no neighbourhoods $U$, $V$ about $x$ and $y$ respectively have trivial intersection. Now consider the point $(x,y)$ that is in the complement of $\Delta$. Now let $U \times V$ be any basis element that contains $(x,y)$ (such an element exists by definition of the product topology being generated by the basis $\mathcal{B}$ consisting of elements of the form $W \times Z$, where $W$ is open in $X$ and $Z$ is open in $Y$). Consider $(U \times V ) \cap \Delta$, which I claim to be $(U \cap X) \times (V \cap X)$.
By our choice of $x$ and $y$ there is $z \in U \cap V$, implying that the intersection $(U \times V ) \cap \Delta$ is not trivial.
Since $U \times V$ was any basis element containing $(x,y)$, this means that $(x,y) \in \overline{\Delta}$, which means that there exists a limit point of $\Delta$ that is not in it, contradicting $\Delta$ being closed.
The problem comes is in the way I have decomposed $\Delta$; the way I have put it seems I am saying that $\Delta$ is equal to $X \times X$, which is not the case. How can I get round this?
Thanks.
Edit: Martin Sleziak has pointed out some mistakes, $(U \times V ) \cap \Delta$ should be $\{ (x,x) : x \in U \cap V\}$ and not as claimed.