Intersection of a closed set and compact set is compact I've been stuck on the following problem for several days: Let $(M,d)$ be an arbitrary metric space and $S, T$ be subsets of $M$.
If $S$ is closed and $T$ is compact, then $S \cap T$ is compact.
I know that if $T$ is compact, $T$ is  closed and bounded. That would imply that $S \cap T$ is also closed and bounded since $(S \cap T) \subseteq T$. Also since $S$ is closed, $S$ contains all its accumulation points.
Other than writing down the definitions, I really don't know how to proceed. Could someone give me a hint?
 A: Hint: Let $\mathcal{C}$ be an open cover of $S\cap T.$ Note that $\mathcal{C}\cup\{M\setminus S\}$ (where $M\setminus S$ denotes the complement of $S$ in $M$) is an open cover of $T$. (Why?) Can you take it from there?
Alternately, as you (astutely) observed, in the comments, you can simply use the Theorem that says a closed subset of a compact metric space is compact, noting that $T$ is a compact metric space and that $S\cap T$ is a closed subset of $T$.
A: To use your definition of compactness, consider an open covering of $S$. Augment it with the complement of $S$, which is open. This gives you an open covering of $T$. Fill the details and finish the proof. 
A: $S\cap T$ is a closed subset of $T$ (relative to $T$, closedness of $T$ is actually not needed). It is a general fact in topology that a closed subset of a compact space is compact. To show that, let $X$ be a compact topological space (or a metric space), $A$ a closed subset of $X$, and $\mathcal U=\{U_i\mid i\in I\}$ an open cover of $A$. Now consider the open cover $\mathcal U \cup\{X-A\}$ of $X$.
