In that case you can't use the Heine-Borel theorem because this theorem only apply to the case of normal topology.
To prove this proposition, you just find the open cover of $[0,1]$ such that every finite subcover does not cover $[0,1]$. Let $\mathcal{A}$ be a set of open sets defined as
$$\mathcal{A}=\{[0,r):0<r<1\}\cup \{[1,2)\}$$
then $\mathcal{A}$ cover $[0,1]$. However every finite subcover of $\mathcal{A}$ does not cover $[0,1]$.
"...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 $\tau_C=\tau_H$.
Best Answer
$[0,1]$ is not connected, e.g. because $[0,\frac12)$ is a non-trivial closed-and-open subset of it.
Not compact as $\{1-\frac1n: n=2,3,4,5,\ldots\}$ is an infinite subset without limit point in it.
Hausdorff because as you state its topology includes the usual, Hausdorff, one.