Attempt at showing intersection of two compact sets is compact using open cover definition and why proof doesn't work? Let $S$ and $T$ both be compact sets in $\mathbb{R}^{n}$. Show that $S \cap T$ is also compact using the open cover definition.
What I thought was going to be a simple decomposition has turned out to be more complicated for me.
Attempt
We want to show that for every open cover $U = \{U_{\alpha}\}$ of $S \cap T$, there exists a finite subcover. 
Suppose there exists an open cover $H$ of $S \cap T$ that has no finite subcover. THis would imply that both $S$ and $T$ have an open cover which has no finite subcover. But it is assumed $S$ and $T$ are compact, so a contradiction.
My Issue
I feel my solution has a big hole in its reasoning and I'm not a fan of contradiction either since I'm not truly working with the concepts. I looked at a few of the solutions posted here and I don't fully understand their approach with regards to this question.  Particularly this solution: Prove Intersection of Two compact sets is compact using open cover?. 
How are we constructing a finite subcover from the compliment?
 A: You may or may not know what it means for a topological space to be Hausdorff but $\mathbb{R}^n$ has this property. It turns out that compact sets are closed in Hausdorff spaces (and hence $\mathbb{R}^n$) and arbitrary intersections of closed sets are also closed (perhaps you are already aware of these facts for subsets of $\mathbb{R}^n$). So, $S, T$ and $S \cap T$ are closed. 
Let $ \mathcal{U} = \{ U_{ \alpha } \}$ be an open cover of $S \cap T$. Then $\mathcal{U} \cup \{ \mathbb{R}^n \setminus (S \cap T) \}$ is an open cover of $S$ as well as for $T$ (why? I leave that to you). So, this cover yields a finite subcover of each $S$ and $T$ (the covers may be distinct but each will cover $S \cap T$). If either finite subcover includes $\mathbb{R}^n \setminus (S \cap T)$ then remove that. What are you left with?
A: As in the comments, this statement actually holds for all Hausdorff spaces - we are not limited to $\mathbb{R}^n$ - and also for arbitrary families of compact sets. But we can work with $\mathbb{R}^n$ here - the proof here can be easily generalised to Hausdorff spaces. 
We let $K_i$ be a family of compact subsets indexed by some arbitrary set $I$. They are closed (you may have this as a characterisation, or by Heine-Borel, but it's also a nice exercise to do, and reasonably straightforward since $\mathbb{R}^n$ is a metric space), and so their intersection $\cap_{i\in I} K_i$ is closed. Then (as another exercise), closed subspaces of compact spaces are themselves compact. Since $\cap_{i\in I} K_i \subseteq K_i$ (for any $i \in I$), the former is compact. 
