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?

According to the definition of the compact set, we need every open cover of set K contains a finite subcover. Hence, not every subsets of compact sets are compact.
Why closed subsets of compact sets are compact?
Proof 
Suppose $F\subset K\subset X$, F is closed in X, and K is compact. Let $\{G_{\alpha}\}$ be an open cover of F. Because F is closed, then $F^{c}$ is open. If we add $F^{c}$ to $\{G_{\alpha}\}$, then we can get a open cover $\Omega$ of K. Because K is compact, $\Omega$ has finite sucollection $\omega$ which covers K, and hence F. If $F^{C}$ is in $\omega$, we can remove it from the subcollection. We have found a finite subcollection of $\{G_{\alpha}\}$ covers F.
