Suppose $T_1$ is a strict subset of $T_2$. I want to show that (X, $T_1$) and (X, $T_2$) cannot be both compact and Hausdorff. Let $X$ be a set and let $T_1$ and $T_2$ be topologies on X. Suppose $T_1$ is a strict subset of $T_2$. I want to show that (X, $T_1$) and (X, $T_2$) cannot be both compact and Hausdorff.
I am stumped by this problem. I tried the following:
If a space is compact and Hausdorff, it is normal. The idea is to show both spaces cannot be normal. So let us take an open set $U$ in $X_2$ that is not contained in $X_1$. Let us take an open set $V$ that is in $X_1$ which contains $U$. Then the set $V-U$ is closed. But I'm not sure how to take this anywhere.
Does anyone else have ideas or hints?

Suppose $(X,T_2)$ is compact and $(X,T_1)$ is Hausdorff.
Define $f: (X, T_2) \rightarrow (X,T_1)$ by $f(x) = x$ (the identity). 
If $O \in T_1$, then $f^{-1}[O] = O \in T_2$ as $T_1 \subseteq T_2$. This shows $f$ is continuous.
Let $U$ be open in $T_2$ but not open in $T_1$ (as we have strict inclusion as a given).
Then $C = X \setminus U$ is closed in $(X,T_2)$, so compact in $(X,T_2)$ as well. So $f[C]$ is compact in $(X, T_1)$ by continuity. As $(X,T_1)$ is Hausdorff (!), $C$ is also closed in $(X,T_1)$. But then $U = X \setminus C$ is open in $(X,T_1)$, which contradicts how $U$ was chosen...
