Prove that if $\displaystyle f: X \rightarrow Y$ is a continuous map of a space X to a Hausdorff space Y then its graph
$\displaystyle G_{f} = ({ (x,f(x)) \in X \times Y : x \in X})$
is a closed subset of X x Y.
Let (x,y) be a point in the closure of $\displaystyle G_f$, and let $\displaystyle \bigl((x_n,f(x_n))\bigr)$ be a sequence of points in $\displaystyle G_f$ with limit point (x,y). Then $\displaystyle x_n\to x$ and $\displaystyle f(x_n)\to y$. But f is continuous, and therefore $\displaystyle f(x_n)\to f(x)$. Therefore y=f(x) and hence $\displaystyle (x,y) = (x,f(x))\in G_f$. Thus every limit point of $\displaystyle G_f$ is in $\displaystyle G_f$, and so $\displaystyle G_f$ is closed.
(If the space is "too big" for closures to be detected by sequences then you will have to replace "sequence" by "directed net" in that proof.)
The proof does use the information that Y is Hausdorff (though I didn't actually point that out). It is used to deduce that if $\displaystyle f(x_n)\to f(x)$ and $\displaystyle f(x_n)\to y$ then $\displaystyle f(x) = y$. (In a Hausdorff space, a sequence has a unique limit. In a non-Hausdorff space, it's possible for a sequence to have more than one limit.)