The product of locally compact spaces is locally compact

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My definition of local compactness is as follows:

A topological space $(X, \mathcal{T})$ is locally compact if, for every $x \in X$, there is a $U \in \mathcal{T}$ with $x \in U$ such that $\overline{U}$ is compact.

I am then asked to prove that, if $X$ and $Y$ are locally compact and Hausdorff, then $X \times Y$ equipped with the product topology is also locally compact. Trouble is, I don't believe I use Hausdorff in my proof! And I don't see where it breaks down. Here it is:

Let $(X, \mathcal{T}_X)$ and $(Y, \mathcal{T}_Y)$ be locally compact. Consider a point $(x, y) \in X \times Y$. Then there exist $U \in \mathcal{T}_X$ with $x \in U$ and $V \in \mathcal{T}_Y$ with $y \in V$ such that $\overline{U}$ is compact in $X$ and $\overline{V}$ is compact in $Y$.

Firstly, $U \times V$ is open in $X \times Y$, and $(x,y) \in U \times V$. Furthermore, $\overline{U}$ is compact as a subset of $X$, so considered as a topological space with the subspace topology, it is also compact. Similarly for $\overline{V}$. Then the space $\overline{U} \times \overline{V}$ equipped with the product topology is also compact. Now I believe that the product topology induced by the topologies on $\overline{U}$ and $\overline{V}$ is the same as the subspace topology induced by considering $\overline{U} \times \overline{V}$ as a subspace of $X \times Y$. So $\overline{U} \times \overline{V}$ is also compact with respect to the subspace topology, and so it is a compact subset of $X \times Y$. But $\overline{U} \times \overline{V} = \overline{U \times V}$, so we're done.

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With that definition of local compactness (there are several, so I'm glad you included it) your proof is indeed correct.

$\overline{U\times V} = \overline{U} \times \overline{V}$ in the product topology and the product of compact sets is compact, in short.

It's very common for people to consider local compactness in the context of Hausdorff spaces only, as then all the different usual definitions are all equivalent to each other, and we can prove many more theorems (and we have a nicely behaved one-point compactification as well). The exercise poser could just have been lazy and didn't realise that for this one result he could do away with the Hausdorffness altogether.