A subset of a compact set is compact? Claim:Let $S\subset T\subset X$ where $X$ is a metric space. If $T$ is compact in $X$ then $S$ is also compact in $X$.
Proof:Given that $T$ is compact in $X$ then any open cover of T, there is a finite open subcover, denote it as $\left \{V_i  \right \}_{i=1}^{N}$. Since $S\subset T\subset \left \{V_i  \right \}_{i=1}^{N}$ so $\left \{V_i  \right \}_{i=1}^{N}$ also covers $S$ and hence $S$ is compact in X
Edited: I see why this is false but in general, why every closed subset of a compact set is compact?

Here's an alternate proof (for closed subsets, obviously): any net on $S$ is a net on $T$ and thus has a convergent subnet with limit is in $T$ -- but its limit must also be in $S$ because $S$ is closed.
It's a little tricky because the notion of closed sets and compact sets are intuitively very similar.
