I want to show that a finite dimensional normed vector space X over $\mathbb{K}$ ( $\mathbb{K}= \mathbb{R}$ or $\mathbb{C}$ ) is always complete by using the fact that X is isomorphic to $\mathbb{K}^n$, which is complete.
We pick a basis $\{ \hat{e_i} \}_{i=1...n}$ for X and the usual basis $ \{ e_i \}_{i=1...n} $ for $\mathbb{K} ^n $ and consider the isomorfism:
$$f: X \longrightarrow \mathbb{K}^n $$ $$\hspace{1cm} \Sigma x_i \hat{e}_i \to \Sigma x_i e_i $$
With no loss of generality we can assume $||\hat{e}_i||=1$, $||e_i||=1$ $\forall i=1...n$.
Now, for every Cauchy sequence $\{ x_i\}_{i \in \mathbb{N}}$ in X, as $f$ is an isometry, $\{ f(x_i)\}_{i \in \mathbb{N}}$ is a Cauchy sequence in $ \mathbb{K}^n $, so $\{ f(x_i)\}_{i \in \mathbb{N}} \to k \in \mathbb{K}^n$.
From the continuity of $f$ and $f^{-1}$ ($f$ and $f^{-1}$ are both linear and bounded), it follows that $\{ f^{-1}(f(x_i))\}_{i \in \mathbb{N}} \to f^{-1}(k) \in X$. As f is bijective, $\{ x_i\} \to x \in X$.
Here's a proof:
We know that in $\mathbb{K} ^n$ all norms are equivalent. Thus, we are going to work with $\| *\|_{1}$. Then, we define
$$f: \mathbb{K}^n \longrightarrow X $$ $$\hspace{1cm} \Sigma x_i e_i \to \Sigma x_i\hat{e}_i$$ You need to see $f$ is an homeomorphism. In order to see this, try to find $L,M>0$ such that
$$ L\|x\|_{1} \leq \|f(x)\| \leq M\|x\|_{1}$$
To Find M use triangle inequality and the definition of $f(x)$.
Now, we know $f$ is continuous. The image of a compact under a continuous function is compact. Take the unit sphere $S_{\mathbb{K}^n}$. $f(S_{\mathbb{K}^n})$ is compact and $0 \notin f(S_{\mathbb{K}^n})$ since $f(0)=0$ and it is bijective. $\{0\}$ is closed. Distance between closed and compact with empty intersection is positive. Let that distance be $L$.
Therefore, $ \|f(x)\|\geq L$ $\forall x \in S_{\mathbb{K}^n}$. Finally, using $y \in \mathbb{K}^n $ , divided by norm, it is proved.