Any closed subset $S$ of a compact space $T$ is compact. I have a lemma which i'm trying to prove:
Any closed subset $S$ of a compact space $T$ is compact.
My proof:
Let $\mathcal{U}$ be a an open cover of $T$
Since $S\subset T \subset \mathcal{U}$, it follows that $S\subset \mathcal{U}$, which is a collection of open sets which union contains $S$ and that is precisely the definition of a cover. Since this cover has a finite subcover, it follows that $S$ is compact.
Would this be correct?
 A: It is not correct. To prove that $S$ is compact you have to start with an open cover of $S$, not an open cover of $T$. Take an open cover of $S$ and include $T\setminus S$ in it and use compactness of $T$ to finish the proof.
A: The first sentence of your proof is incorrect.
You need to prove that $S$ is compact. To prove that $S$ is compact, you must prove that any open cover of $S$ has a finite subcover. In order to prove that, the first sentence of your proof must be

Let $\mathcal U$ be an open cover of $S$.

instead of what you wrote.

What you sort of proved is that every open cover of $T$ contains an open subcover of $S$, which is not the same.

Also, note that you also incorrectly wrote that $T\subset \mathcal U$. This relation is incorrect. If $\mathcal U$ is a cover of $T$, then the elements of $\mathcal U$ are subsets of $T$. So, you can write that $$T\subset \bigcup_{U\in\mathcal U} U ,$$
but you cannot write $T\subset \mathcal U$.
A: You have to take an open cover of $S$ not of $T$ because you like to show, that $S$ is compact. Take an open cover $\mathcal U$ of $S$. Then add $T\setminus S$, which is open, since $S$ is closed. Now you have an open cover of $T$ with a finite subcover, because $T$ is compact. Now substract $T\setminus S$ and you get a finite subcover of the beginning cover of $S$.
