The Heine-Borel theorem for $\mathbb{R}^n$ states that a subset $K\subset\mathbb{R}^n$ is closed and bounded if and only if it is compact. (A set is said to be compact if every open cover of the set has a finite sub-cover). The standard method to show that closed- and boundedness implies compactness starts by enclosing $K$ in a box, and continually bisecting it until some or other contradiction is obtained. I have an alternative approach, but I am not entirely sure it works, and I would like to know if it seems possible to extend this to an arbitrary metric space (by replacing closed and bounded by sequentially compact) beyond that, I think its interesting that an inductive proof for it exists, so here it is:
Firstly, it is sufficient to prove that every closed ball centered at $0$ is compact, since one can always enclose a bounded set $K$ inside such a closed ball of radius $r$, $U(r)$, and append to any cover of $K$ the set $\mathbb{R}^n \setminus K$, and finding a finite subcover of the new cover is then equivalent to finding a finite subcover of the original cover.
The base case is trivial since $\mathbb{R}^{0}=\{0\}$ has only compact subsets. If we then assume that this theorem holds in $\mathbb{R}^k$, we will show that it holds in $\mathbb{R}^{k+1}$.
Now let $U(r)\subseteq \mathbb{R}^{k+1}$ be the closed ball of radius $r$ centered at the origin, and let $G$ be a cover for $U(r)$. Then we say that $l\in \mathbb{R}^+$ is reached if there is some finite subset $G'$ of $G$ such that
$$U(l) \subseteq \bigcup G' $$
Now define $w$ as supremum of those $l\leq r$ which are reached. Now we assume that $w<r$ and obtain a contradiction. Note that since
$$\partial U(w) \subseteq U(r) \subseteq \bigcup G $$
We have that $\partial U(w)$ is covered by $G$. But $\partial U(w)$ is a $k$-dimensional sphere, so it's upper (inclusive) hemisphere is homeomorphic to the closed ball in $\mathbb{R}^k$, which is compact by assumption, so there is a finite subset of $G$ which covers the upper hemisphere of $\partial U(w) $, the same is of course true for the lower hemisphere, so there is a finite subset of $G$, call it $G_0$, which covers $\partial U(w) $. It is then easy to show that $G_0\cup G'$, where $G'$ is a finite subset of $G$, covers an open ball with a radius larger than $w$ (the proof for this is included at the end) and thus it covers a closed ball with a radius larger than $w$, which is a contradiction, since $w$ was assumed to be the supremum of reached radii.
Some of my questions are, is this a known proof? Is it correct?
Proof that an open ball of radius larger than $w$ is covered by a finite subset of $G$:
To see why, it is sufficient to show that there is some $\delta>0$ such that for any $x\in\partial U(w)$, the open ball of radius $\delta$ is contained in some member of $G_0$. Suppose this is not true, then for any $\delta>0$ there is some $x\in\partial U(w)$ such that the open ball of radius $\delta$ is not contained in any member of $G_0$, we can in particular take the convergent sequence of $\delta_n=\frac{1}{n}$, which induces a sequence $(x_n)$ with elements in $\partial U(w)$, but because of sequential compactness, this sequence converges to some point$x_0\in\partial U(w)$ but this is a contradiction since $x_0$ is contained in some open ball of radius $\delta_0$ (because $G_0$ is an open cover for $\partial U(w)$) and so eventually the $x_n$'s will be within $\frac{\delta_0}{2}$ of $x_0$, but then each $x_n$ beyond this point is contained in an open ball of radius $\frac{\delta_0}{2}$, a contradiction, so there is some $\delta>0$ such that every point in $\partial U(w)$ is contained in the open ball of radius $\delta$ around that point. A routine application of the triangle inequality then shows that for any $|\varepsilon|\leq \frac{\delta}{2}$ we have
$$\partial U(w+\varepsilon) \in \bigcup G_0$$
Since there is some finite subset $G'\subseteq G$ such that
$$U\left(w-\frac{\delta}{2}\right)\subseteq \bigcup G' $$
So $G'\cup G_0$ covers $U(w+\frac{\delta}{2})$ which is the contradiction we required in the original proof
Best Answer
You correctly state
Your alternative proof is absolutely correct. Its essential part is Proof that an open ball of radius larger than $w$ is covered by a finite subset of $G$. This is based on the fact that in metric space compactness is equivalent to sequential compactness. Although this is true, it seems to be a bit unsatisfactory because you invoke an argument of a new provenance ( not an open-cover-argument). We shall see later that this is essentially a variant of the tube lemma.
The tube lemma is easily proved by considering open covers. It is used to prove (inductively) that finite products of compact spaces are compact.
A variant of your proof is to show that that cubes $[-r,r]^n$ are compact. Step 1 is to prove that $[-r,r]$ is compact, step 2 is to use induction.
There are various proofs of the compactness of closed intervals $[a,b]$, and one of these proofs uses your "reached" argument which is very easy in this case. See How to prove every closed interval in R is compact?
Let us modify your proof and avoid the use of a sequential compactness argument.
$\partial U(w)$ is a $k$-dimensional sphere, and you correctly argue that it is compact by assumption (it is the union of upper and lower hemisphere which are homeomorphic to the closed ball in $\mathbb R^k$). Thus $\partial U(w)$ is covered by a finite subset $G_0$ of $G$. The union of the elements in $G_0$ is an open neighborhood $U$ of $\partial U(w)$. We wish to show that $U$ contains a spherical shell $S(w,\delta) = \{ x \in \mathbb R^{k+1} \mid w -\delta < \lVert x \rVert < w + \delta \}$. If we know this, we pick $l$ such that $w -\delta < l < w$. Then $U(l)$ is covered by a finite subset $G_l$ of $G$. The set $G' = G_0 \cup G_l$ is finite and covers $U(w+\delta)$ which contradicts the definition of $w$.
How to find $\delta$? Consider the homeomorphism $$h : \mathbb R^{k+1} \setminus \{0\} \to S^k \times (0,\infty), h(x) = \left(\frac{x}{\lVert x \rVert}, \lVert x \rVert\right)$$ whose inverse is given by $$h^{-1}(y,t) = ty.$$ Then $h(U)$ is an open neighborhood of $S^k \times \{w\}$. By the tube lemma there is $\delta > 0$ such that $V = S^k \times (w-\delta, w+\delta) \subset h(U)$. Then $h^{-1}(V) \subset U$ is the desired spherical shell $S(w,\delta)$.