If $K\subset{\mathbb{R}}$ closed and bounded, then K compact Proof (without covers)

Question

I have already proved the reverse (Heine-Borel Theorem) by use of the Bolzano Weierstrass Theorem, but, reading several proofs of this forward direction, am yet to come across a (short) proof which shows this without the need for open covers - note, here I define a compact set to be a subset $K\subset{\mathbb{R}}$ such that every sequence in $K$ has a subsequence that converges to a limit that is also in $K$ Could someone please show thus this forwards direction (closed and bounded, then compact) without resorting to open covers (not for lack of knowledge, but as covers appear on the next page of this theorem in the book I'm studying).

Answer 1

Let $(x_n)_{n \geq 1} \subseteq K$. Since $K$ is bounded, so is this sequence. By Bolzano-Weierstraß, there exists a convergent (in the whole space) subsequence $(x_{n_j})_{j \geq 1}$. Since $K$ is closed, this subsequence converges in $K$.

Answer 2

Let $ x_n $ be a sequence in $ K $, which is a closed and bounded subset of $ \mathbb R $. It is a well known property of $ \mathbb R $ that any sequence has a monotone subsequence, so let $ x_{k_n} $ be a monotone subsequence. Then $ x_{k_n} $ is a monotone, bounded (as $ K $ is bounded) sequence in $ \mathbb R $. By the least upper bound property of $ \mathbb R $, $ x_{k_n} $ converges to some $ x \in \mathbb R $. Then as $ K $ is closed and $ x_n $ is a sequence in $ K $, $ x \in K $, which shows compactness of $ K $.