Rudin's Proof of Bolzano-Weierstrass theorem

Question

(a) If $\{P_n\}$ is a sequence in a compact metric space $X$, then some sub­sequence of $\{P_n\}$ converges to a point of $X$.
(b) Every bounded sequence in $\mathbb R^k$ contains a convergent subsequence.

Rudin proves (a), then argues for (b) as follows:

"(b) This follows from (a), since Theorem 2.41 implies that every bounded subset of $\mathbb R^k$ lies in a compact subset of $\mathbb R^k$," where Theorem 2.41 is the Heine-Borel Theorem.

But doesn't this require the set to be bounded AND closed? From what I can see Rudin makes no argument about that the sequence in $\mathbb R^k$ is closed.

Answer 1

He is not saying that the bounded set itself is compact. What he is saying is that that bounded set is contained in some compact set. As a bounded set is, by definition, contained in an open ball, it is also contained in the closure of the ball, which is compact.

Answer 2

Note than (a) asks only that the sequence is "in a compact metric space". So it doesn't require the sequence to be a closed set, only that it be contained in a compact set. The points in $\mathbb R^k$ with norm $\le M$ are a closed and bounded (ergo compact) subset of $\mathbb R^k$.