Any finite dimensional normed linear space over a complete field is complete. 
Prove that any finite dimensional normed linear space over a complete field is complete.

After several comments and corrections, I present the correct proof in the answer section.
 A: PROOF
Let $E$ be any finite dimensional normed linear space over a complete field, $\Bbb{R}$ or $\Bbb{C},$ say. Suppose $\dim E=n\geq 1,$ and let $\{e_i\}^{n}_{i=1}$ be a basis for $E.$ Then, there exists scalars $\{\alpha_i\}^{n}_{i=1}$ such that, for arbitrary $x\in E$,
\begin{align} x= \sum^{n}_{i=1} \alpha_i e_i .\end{align}
Suppose $\{x_r\}_{r\in \Bbb{N}}$ is Cauchy in $E$ w.r.t $\|\cdot\|$ norm and $\epsilon'>0.$ Then, there exists $N$ such that for all $s\geq r\geq N,$
\begin{align} \|x_r-x_s\|<\epsilon'.\end{align}
Now, $\|\cdot\|_1$ defined by $\|x\|_1=\sum^{n}_{i=1} |\alpha_i|$ is a norm on $E$ and since All norms defined on a finite dimensional normed linear space are equivalent, we have that $\|\cdot\|_1\sim \|\cdot\|,$ i.e., there exists $\gamma,\beta>0,$ such that
\begin{align} \gamma\|x\|_1\leq\|x\|\leq \beta\|x\|_1,\;\forall\,x\in E.\end{align}
Then,
\begin{align}\gamma|\alpha_{i,r}-\alpha_{i,s}|\leq\gamma\sum^{n}_{i=1}|\alpha_{i,r}-\alpha_{i,s}|=\gamma\|x_r-x_s\|_1\leq\|x_r-x_s\|<\epsilon',\;s\geq r\geq N\end{align}
and so the sequence $\{\alpha_{i,r}\}^{n}_{i=1,r\in \Bbb{N}}$ is Cauchy in $\Bbb{R}$ or $\Bbb{C}$. Now, $\alpha_{i,r}\to \alpha_{i}\in \Bbb{R}\;\text{or}\;\Bbb{C},\;\text{as}\;r\to \infty$ by completeness. This implies that for each $i\in\{1,2,\cdots,n\},$ there exists $N_i:=N(i,\epsilon')$ such that 
\begin{align}|\alpha_{i,r}-\alpha_{i}|<\epsilon',\;r\geq N_i.\end{align}
Taking $M=\max\{N_i:1\leq i\leq n\},$ we have that
\begin{align}|\alpha_{i,r}-\alpha_{i}|<\epsilon',\;r\geq M.\end{align}
Let $\epsilon>0$ and $n\in\Bbb{N}$, then for $\epsilon'=\dfrac{\epsilon}{n\beta},$ there exists $M$ such that \begin{align}|\alpha_{i}-\alpha_{i,s}|\leq\dfrac{\epsilon}{n\beta},\;s\geq M.\end{align}
Taking sums, we have
\begin{align}\|x-x_s\|\leq\beta\|x-x_s\|_1=\beta\sum^{n}_{i=1}|\alpha_{i}-\alpha_{i,s}|\leq\sum^{n}_{i=1}\dfrac{\epsilon}{n},\;s\geq M.\end{align}
Hence, \begin{align}\|x_s-x\|\leq\epsilon,\;s\geq M,\end{align}
and so, we have that $x\in E,$ which implies that $E$ is a complete finite dimensional normed linear space and we are done!
