Proof of part of the Heine-Borel Theorem I wanna prove a portion of the Heine-Borel Theorem because I didn't like the proof I was given. I'm only going to be proving the closed part. I will not be proving the bounded part because that part was fine.
Theorem In $\mathbb{R}^n$ with the Euclidean topology, compact sets are precisely the closed and bounded subsets of $\mathbb{R}^n$
Proof
Let $K$ be a compact (bounded) subset of $\mathbb{R}^n$. Toward a contradiction, suppose $K$ is not closed.
Since $K$ is not closed, let $x_o$ be an accumulation point of $K$ which is not in $K$.
Consider all $k \in K.$ Since $K$ is bounded, let $M =$ max$\{|k_1|,|k_2|,|k_3|,...\}$. Then for all $k \in K,$ $-M \leq k \leq M,$ bounding all of $K$.
Consider $\mathcal{U}_i =\{k \in K: d(k,x_o)<\frac{M}{i}\}$ for $i \in \mathbb{N}$
Then $\displaystyle\bigcup_{i=1}^\infty \mathcal{U}_i$ is an open cover of $K$ with no finite subcover
This contradicts our assumption that $K$ was not closed.
Thus, the compact set $K$ is closed.
 A: $M$ as written is not well-defined. If $K$ is infinite, then the set $\{|k_1|, |k_2|, \dots\}$ may not have a maximum element.
There is a type error in your inequality $-M \leq k \leq M$. $M$ is a real number, but $k$ is a vector.
Similarly, there is a type error in "$\displaystyle\bigcup_{i=1}^\infty \mathcal{U}_i$ is an open cover...". Open covers are sets of subsets, $\displaystyle\bigcup_{i=1}^\infty \mathcal{U}_i$ is just a single set.
Furthermore, the most important part, that, $\mathcal{U}_i$ is an open cover without a finite subcover is, well, wrong. Each $\mathcal{U}_i$ is smaller than the one proceding it, so a finite subcover would be given by $\mathcal{U}_1$.
A: Duncan Ramage has explained what is wrong with your argument. An argument along somewhat similar lines is possible, however, and is actually a bit simpler than what you were trying to do, since it does not require any information on whether $K$ is bounded. It begins as you did.
If $K$ is not closed, let $x_0\in\operatorname{cl}K\setminus K$, for $k\in\Bbb Z^+$ let
$$U_k=\left\{x\in\Bbb R^n:d(x,x_0)>\frac1k\right\}\,,$$
and let $\mathscr{U}=\{U_k:k\in\Bbb Z^+\}$; then $\bigcup\mathscr{U}=\bigcup_{k\ge 1}U_k=\Bbb R^n\setminus\{x_0\}$, so $\mathscr{U}$ is an open cover of $K$.
Suppose that $\{U_{k_1},\ldots,U_{k_m}\}$ is a finite subset of $\mathscr{U}$, where $k_1<\ldots<k_m$. Then $U_{k_1}\subseteq\ldots\subseteq U_{k_m}$, so $\bigcup_{i=1}^mU_{k_i}=U_{k_m}$. Since $x_0\in\operatorname{cl}K$, there is a point $x\in K$ such that $d(x_0,x)<\frac1{k_m}$; clearly
$$x\in K\setminus U_{k_m}=K\setminus\bigcup_{i=1}^mU_{k_i}\,,$$
so $\{U_{k_1},\ldots,U_{k_m}\}$ does not cover $K$. Thus, no finite subset of $\mathscr{U}$ covers $K$, and $K$ is therefore not compact.
