Show that $\mathbb{R^{n}}$ is not compact by definition I want to show by definition that $\mathbb{R^{n}}$ is not compact, so I did this:
Let ${\Omega}$ be an open cover of $\mathbb{R^n}$ and suppose that $\mathbb{R^n}$ is compact, then there exists a finite subcollection of ${\Omega}_{i=1}^{k}$ such that $\mathbb{R^{n}}$ is covered by $\Omega_{i=1}^{k}$. So, ${\cup}_{i=1}^{k}\Omega_{i}$ $\supseteq \mathbb{R^{n}}$.
From this part, I don't know how to continue the proof, I think the last argument is a contradiction but I'm not quite sure. Can you help me? Please.

How about the open cover consisting of the open balls of radius $1$:  $\mathscr C=\{B(r,1)\mid r\in\Bbb R^n\}$?
I don't think there's a finite subcover.
For suppose we have a finite subcollection:  $\mathscr F=\{B(r_k,1)\mid k=1,\dots,n\}$.  
Take a point $x\in\Bbb R^n$ with distance from the origin $d(x,0)\gt\max \{1+d(r_k,0)\mid k=1,\dots,n\}$.  You can do this because $\Bbb R^n$ is unbounded.   (I guess technically you need the Archimedean principle for this.)
Then $x\not\in B(r_k,1)$ for any $B(r_k,1)\in\mathscr F$.
