To prove Heine-Borel theorem for $\mathbb R^n$ with usual Euclidean topology To prove that any closed and bounded subset of $\mathbb R^n$ is compact , I proceed as : Since $\mathbb R^n$ is complete so any closed subset of it is complete . Then I show that any bounded subset of $\mathbb R^n$ is totally bounded , then use complete and totally bounded implies compact . Are there any other approaches  ? Is there any approach which is somewhat topological ? Thanks in advance 
 A: I will proof this for $n=1$, so for $\mathbb{R}$. The general case similar but it needs more attention to details.
Let $A \subset \mathbb{R}$ be totally bounded and closed.
We want to prove that $A$ is compact.
Proof: 
We know that $A$ is totally bounded. So: $$\exists M>0: A \subset [-M,M]$$
We show that $[-M,M]$ is compact. So the closed subset $A$ of this is compact to.
Let $$[-M,M] \subset \mathop{\bigcup_{\alpha \in I}} {G}_{\alpha}$$
be an infinite open covering of $M$.
Denote $S_{0} = [-M,M]$. Assume that $S_{0}$ can not be covered by finite elements of $I$. Then at least one of $[-M,0]$, $[0,M]$ can not be covered by finite elements from $I$. Call this interval $S_{1}$. Notice: $S_{1} \subset S_{0}$.
By extending this, we have constructed a chain of sets:
$$S_{0} \supset S_{1} \supset ...\supset S_{n} \supset ...$$
And each set of this chain can not be covered by finite elements of $I$.
Now $d(S_{n})=\frac{2M}{2^{n}} \rightarrow 0$.
Because each $S_{n}$ is closed, there exists an element $x \in \mathbb{R}$: (See below why)
$$\mathop{\bigcap_{n \in \mathbb{N}}{S_{n}}} = \{x\}$$
We know: $$x \in S_{0}=[-M,M] \subset \mathop{\bigcup_{\alpha \in I}} {G}_{\alpha}$$ So:
$$\exists \alpha \in I: x \in G_{\alpha}$$
$ G_{\alpha}$ is open, so: $\exists \epsilon>0: B(x,\epsilon) \subset G_{\alpha}$. Because $d(S_{n}) \rightarrow 0$, we can take $N \in \mathbb{N}$ big enough so that: $S_{N} \subset G_{\alpha}$.
We see that $S_{N}$ is covered by only one element of $I$. This is in contradiction with the assumption that $S_{N}$ could not be covered by finite elements of $I$. So $[-M,M]$ can be covered by finite elements of $I$. So $[-M,M]$ is compact.
This concludes the proof
Im gonna proof the 'see below why' step.
Let $$S_{0} \supset S_{1} \supset ...\supset S_{n} \supset ...$$
be a chain of closed intervals in $\mathbb{R}$ with $d(S_{n}) \rightarrow 0$.
Then $$\exists x \in \mathbb{R}: \mathop{\bigcap_{n \in \mathbb{N}}{S_{n}}} = \{x\}$$
proof
Denote $[a_{n},b_{n}] = S_{n}$
We directly see that $(a_{n})$ is an increasing sequence in $\mathbb{R}$ and $(b_{n})$ is a decreasing sequence in $\mathbb{R}$ with: $a_{n} \leq b_{n}$ so these are convergent sequences. Denote: $$\mathop{\lim_{n \rightarrow \infty}{a_{n}}} = a$$
$$\mathop{\lim_{n \rightarrow \infty}{b_{n}}} = b$$
trivially: $S_{n} \supset [a,b]$ for each $n \in \mathbb{N}$. So $$\forall n \in
\mathbb{N}: b-a \leq d(S_{n})$$
Taking the limit of $n$ approaching $\infty$ we have: $b-a = 0$. So $a=b$
We also see: $$a \in \mathop{\bigcap_{n \in \mathbb{N}}{S_{n}}}$$
Let $$c \in \mathop{\bigcap_{n \in \mathbb{N}}{S_{n}}}$$. Then for each $n \in \mathbb{N}$, we have: $c \in S_{n}$. By a similar argument as earlier:
$$ \forall n \in \mathbb{N}: |c-a| \leq d(S_{n})$$
So $c = a$.
$$\mathop{\bigcap_{n \in \mathbb{N}}{S_{n}}} = \{ a \}$$
