"...every subspace of compact Hausdorff is closed."
I'm not sure what you mean by this. The space $[0,1]$ is compact and Hausdorff, but its subspace $[0,1/2)$ is not closed.
"...every closed set under the topology $\mathcal{T}$ can be written as finite union of closed sets under the topology $\mathcal{T}'$."
Well if $E$ is $\mathcal{T}$-closed, then $E^c$ is $\mathcal{T}$-open, and thus by $\mathcal{T}\subseteq\mathcal{T}'$, we see that $E^c$ is $\mathcal{T}'$-open and consequently $E$ is $\mathcal{T}'$-closed. Thus every $\mathcal{T}$-closed set is $\mathcal{T}'$-closed. However how does this show that $\mathcal{T}\supseteq\mathcal{T}'$?
The easiest way to do this is to assume $\mathcal{T}\subseteq\mathcal{T}'$, and consider the identity map $f:(X,\mathcal{T}')\to(X,\mathcal{T})$ defined by $f(x)=x$. This is continuous by our assumption, and it's a bijection. Since it maps from a compact space into a Hausdorff space, we can conclude that it is a homeomorphism. Therefore $f$ maps open sets to open sets, meaning that if $U$ is $\mathcal{T}'$-open, then $U=f(U)$ is $\mathcal{T}$-open. Therefore $\mathcal{T}=\mathcal{T}'$.
If you aren't familiar with the theorem I used above, you should try proving it independently. It is tremendously useful and its proof relies on the same techniques one would employ to prove your problem without it. Here it is:
If $f:X\to Y$ is a continuous function from a compact space to a Hausdorff space, then $f$ is a closed map (i.e., it maps closed subsets of $X$ to closed subsets of $Y$). Moreover, if $f$ is a bijection, then $f$ is a homeomorphism.
I alluded that you can prove that $\mathcal{T}\subseteq\mathcal{T}'$ implies $\mathcal{T}=\mathcal{T}'$ without using this theorem, and as I suspect this is the route you intended, here's how one would do so.
Suppose $U$ is $\mathcal{T}'$-open. Then $U^c$ is $\mathcal{T}'$-closed. Since closed subsets of compact sets are compact, we know that $U^c$ is $\mathcal{T}'$-compact. As you pointed out, $\mathcal{T}'$-compactness implies $\mathcal{T}$-compactness, so $U^c$ is $\mathcal{T}$-compact. Since $\mathcal{T}$ is Hausdorff and compact subsets of Hausdorff spaces are closed, we deduce $U^c$ is $\mathcal{T}$-closed. Therefore $U$ is $\mathcal{T}$-open, which completes the proof.
I find that the following reformulation of this problem is more enlightening.
Let $\tau_C$, $\tau_H$, and $\tau$ be topologies on a set $X$ such that $(X,\tau_C)$ is compact and $(X,\tau_H)$ is Hausdorff.
(i) If $\tau\subseteq\tau_C$, then $(X,\tau)$ is compact.
(ii) If $\tau\supseteq\tau_H$, then $(X,\tau)$ is Hausdorff.
(iii) If $\tau_H \subseteq \tau_C$, then $\tau_C=\tau_H$.
Not by contradiction but by contrapositive:
Let $\mathcal{U}$ be an open cover for $X$ (which satisfies the FIP property).
Then $\mathcal{F} = \{X\setminus U: U \in \mathcal{U}\}$ is a family of closed sets such that $\bigcap \mathcal{F} = X \setminus \bigcup \{U : U \in \mathcal{U}\} =\emptyset$, by de Morgan. This means that $\mathcal{F}$ does not have the FIP.
So there finitely many $F_1 =X\setminus U_1, \ldots F_n = X\setminus U_n$ from $\mathcal{F}$ that have empty intersection. The $U_1,\ldots,U_n$ then form a finite subcover for $\mathcal{U}$ as required, again by de Morgan.
Best Answer
If $X$ is considered with the subspace topology, the inclusion $i:X\to Y$ is continuous. Hence $i(C)\subset Y$ is compact as well.