If (X,T$_1$), (X,T$_2$) are compact and Hausdorff for T$_1$ and T$_2$ which are comparable prove T$_1$ = T$_2$.
Well my idea was to create a function F between (X,T$_1$) and (X,T$_2$) that carry one set to his identity, so I take C a closed set from (X,T$_1$) and as the working space is compact we can affirm that C is compact (so it has a finite open cover by elements of T$_1$).
As the identity function is continuous F(C)$\subset$(X,T$_2$) is a compact set in (X,T$_2$) and due to this we can affirm that T$_1$$\subset$T$_2$.
To prove T$_2$$\subset$T$_1$ is the same idea.
I don't know if my idea is correct or not.
If $\mathcal{T}_1 \subseteq \mathcal{T}_2$ (WLOG), consider the identity map $1_X(x)=x$
$$1_X: (X, \mathcal{T}_2) \to (X,\mathcal{T}_1)$$
For $O \in \mathcal{T}_1$ we have that $1_X^{-1}[O] = O \in \mathcal{T}_2$ so that $1_X$ is continuous.
If $C$ is closed in $\mathcal{T}_2$, then $C$ is compact because a closed subspace of a compact space is compact and then $1_X[C]=C$ is compact in $(X,\mathcal{T}_1)$ too, and as that space is Hausdorff, $C$ is closed in $(X,\mathcal{T}_2)$ and so $1_X$ is a closed and continuous bijection and this means that $1_X$ is a homeomorphism and this implies $\mathcal{T}_1 = \mathcal{T}_2$.