Proof $\mathbb{R}^n$ is a complete metric space. 
$\mathbb{R}^n$ is a complete metric space.

Consider a Cauchy sequence $\{\mathbf{x}_k\}$ in $\mathbb{R}^n$, we want to show it converges to a point $\mathbf{x} \in \mathbb{R}^n$. That is to say, if $|\mathbf{x - x_k}| \to 0$ as $k \to \infty$.
Hence, we let $\epsilon \to 0$, and we get $\mathbf{|x_k - x_j|} < \epsilon$ by Cauchy sequence, and let $\mathbf{x = x_j}$ we showed the desired result.

Definition $\mathbb{R}^n$ is a complete metric space. Every Cauchy sequence in $\mathbb{R}^n$ converges to a point of $\mathbb{R}^n$.
Definition Cauchy sequence. Given $\epsilon > 0$, there is an integer $K$ such that $\mathbf{|x_k - x_j|} < \epsilon$ for all $k,j \geq K$.


I am not fond of my proof, because I am not certain if I can approach $\epsilon$ to be zero, nor if I can equate $\mathbf{x}$ to be $\mathbf{x_j}$ since $\mathbf{x_j}$ is changing while $\epsilon$ changes.
Edit Especially, I am baffled that why we need to do it in coordinates? I think they can be subtracted directly, as the definition of Cauchy sequence I added a short while ago?
 A: Every Cauchy sequence is bounded, hence contained in a compact cube, admits thus a convergent subsequence and therefore converges itself.
A: This proof isn't quite right.
How did we prove that Cauchy sequences converge in $\mathbb R$?  In fact, we didn't, this is just supposed to follow from the construction of the real numbers. Try to see the shortcomings of your method in this context. Or, note that any proof of completeness should fail on $(0,1)$ with the sequence $x_k=\frac1k$, and see why your method doesn't raise any red flags where it should.
How can we use the completeness of $\mathbb R$ to deduce the completeness of $\mathbb R^n$?  
Hint: try considering the sequence that you get by looking at a particular coordinate of $\mathbf x_k$. Why is this sequence Cauchy, and how does that help?
Answer: given $\mathbf x_j\in\mathbb R^n$, let $x_{i,j}$ be the $i^{th}$ coordinate of $\mathbf x_j$.  For each $i$, $\{x_{i,j}\}_{j\geq1}$ is a Cauchy sequence in $\mathbb R$ (Because if one component does not converge, the norm does not.), and thus converges in $\mathbb R$.  Let $x_i$ be the limit of this sequence.  
Now, let $\mathbf x=(x_1,x_2,\dots,x_n)$ with $x_i$ as defined above.  
Consider any $\epsilon>0$. By the convergence of the coordinate sequences, we may select an integer $K$ so that for each $j>K:|x_{i,j}-x_i|<\epsilon/n$.  We note that for $j>K:$
$$
\|\mathbf x_j-\mathbf x\|=\\
\|((x_{1,j}-x_1),(x_{2,j}-x_2),\dots,(x_{n,j}-x_n))\| \leq\\
|x_{1,j}-x_1|+|x_{2,j}-x_2|+\dots+|x_{n,j}-x_n)| <\\
\epsilon/n + \epsilon/n + \dots + \epsilon/n = \epsilon
$$
Thus, $\mathbf x_j\to\mathbf x$, which means that an arbitrary Cauchy sequence must converge.
