analysis - Show that $Bbb{R^n}$ with the $ell^2$-norm is complete

I want to show that $\Bbb{R^n}$ where $\Vert x \Vert_{R^n}=\left(\sum_{i=1}^{n}\left| x_i \right| ^2\right)^{1/2}$ is complete

Here is what I've done.

Let $\{x^{(s)}\}\subseteq (R^n,\Vert \cdot \Vert_{R^n})$ be a Cauchy sequence and $\epsilon>0.$ Then, $\exists\, N\in \Bbb{N}$ s.t. $\forall r\geq s\geq N,$
\begin{align}\Vert x^{(r)}-x^{(s)} \Vert_{R^n}=\left(\sum_{i=1}^{n}\left| x_{i}^{(r)}-x_{i}^{(s)} \right|^2 \right)<\epsilon^{2}.\end{align}
Hence, we have that $x_{i}^{(r)}\to x_{i}^{*}\in \Bbb{R},\;\text{as}\;r\to\infty$, since $\Bbb{R}$ is complete.

Fix $n,r\in \Bbb{N}$, then allow $t\to\infty.$ We have
\begin{align}\left(\sum_{i=1}^{n}\left| x_{i}^{(r)}-x_{i}^{*} \right|^2 \right)<\epsilon^{2},\;\;\forall \;r\geq N, n\in \Bbb{N}.\end{align}

For $r=N,$
\begin{align}\left(\sum_{i=1}^{n}\left| x_{i}^{(N)}-x_{i}^{*} \right| ^2\right)<\epsilon^{2},\;\;\forall \; n\in \Bbb{N}.\end{align}
Hence, $x^{N}-x^{*}\in (R^n,\Vert \cdot \Vert_{R^n})$ and since $(R^n,\Vert \cdot \Vert_{R^n})$ is a linear vector space, then $x^{*}=x^{N}-(x^{N}-x^{*})\in (R^n,\Vert \cdot \Vert_{R^n}),$ and we are done!

Kindly check if I'm correct. Corrections and alternative proofs are welcome.

Answer

Looks good to me.

By the way, since each $x^*_i\in\Bbb R$ we already have $x^*=(x^*_1,\dots,x^*_n)\in\Bbb R^n$ by definition. You don't need that fact that $\Bbb R^n$ is a linear space to conclude that.