Let $(V,\vert\vert\cdot\vert\vert)$ be a Banach space. Prove if $W\subset V$ and $\dim(W)=n$ then $W$ is a closed subset of $V.$ The original (with words) problem statement is:
Let $(V,\vert\vert\cdot\vert\vert)$ be a Banach space. If $W$ is a finite dimensional subspace of $V$ then $W$ is a closed subset of $(V,\vert\vert\cdot\vert\vert).$
My attempt at the proof is as follows:
Since $W$ is finite dimensional suppose that $\dim(W)=n$ and let $\{w_1,\dots,w_n\}$ be a basis for $W$. Define $\varphi:W\to\mathbb{R}^n$ as $\varphi(c_1w_1+\cdots+c_nw_n)=(c_1,\dots,c_n)$, which in fact defines an isomorphism between $\mathbb{R}^n$ and $W$. To show that $W$ is closed let $(w_k)_{k=1}^\infty$ be a sequence of vectors taken from $W$ that converges to $w$. We want to show that $w\in W$. 
Now, for every $k\in\mathbb{N}$ we have that $w_k=c_1^{(k)}w_1+\dots+c_n^{(k)}w_n$.  And so we have that $\varphi(w_k)=\varphi(c_1^{(k)}w_1+\cdots+c_n^{(k)}w_n)=(c_1^{(k)},\dots,c_n^{(k)}).$ Now since $w_k\to w$, where $w=c_1w_1+\dots+c_nw_n$ and that $\varphi$ is an isomorphism, we have that $\lim\limits_{k\to\infty}w_k=\lim\limits_{k\to\infty}\varphi^{-1}(c_1^{(k)},\dots,c_n^{(k)})=\varphi^{-1}(c_1,\dots,c_n)=w$. Since $\mathbb{R}^n$ is complete, $(c_1,\dots,c_n)\in\mathbb{R}^n$, which means that $\varphi^{-1}(c_1,\cdots,c_n)=w$ is in $W$. Since $(w_k)_{k=1}^\infty$ was an arbitrary sequence taken from $W$ which has the limiting vector living in $W$, $W$ must be closed.
I think what I have is right, but I would very much appreciate any type of feedback to improve on my proof or note any mistakes I've made. 
 A: Here is a proof that any two norms on a finite dimensional vector space are equivalent.
Step 1 (Settings). Let $(V, \| \cdot \|)$ be a normed vector space of dimension $\dim V = n$. Choose any basis $\{ u_{1}, \cdots, u_{n} \}$ of $V$ and define the norm $\| \cdot \|_{2}$ on $V$ by
$$ \| a_{1}u_{1} + \cdots + a_{n}u_{n} \|_{2} = (a_{1}^{2} + \cdots + a_{n}^{2})^{1/2}. $$
In other words, we identify $V$ with $\Bbb{R}^{n}$ as normed spaces using the coordinate map. This identification shows that $\| \cdot \|_{2}$ is indeed a norm on $V$, and the unit sphere
$$B = \{ v \in V : \| v \|_{2} = 1 \}$$
is compact w.r.t. $\| \cdot \|_{2}$. (This is the crucial property that makes this whole proof work.)
Step 2. The claim follows once we check that $\| \cdot \|$ and $\| \cdot \|_{2}$ are equivalent: there exist two constants $c, C > 0$ such that
$$ c\| v\|_{2} \leq \| v\| \leq C \| v \|_{2}. \tag{*} $$
The second part of $\text{(*)}$ is easy to prove. Let $C = \|u_{1}\| + \cdots \|u_{n}\|$. Then
\begin{align*}
\| v\|
 = \| a_{1}u_{1} + \cdots + a_{n}u_{n} \|
&\leq |a_{1}| \|u_{1}\| + \cdots + |a_{n}| \|u_{n}\| \\
&\leq \|v\|_{2} \|u_{1}\| + \cdots + \|v\|_{2} \|u_{n}\|
= C\|v\|_{2}.
\end{align*}
Step 3. To prove the first part of $\text{(*)}$, assume otherwise. Then there exists $v_{k} \in V - \{0\}$ such that 
$$ \frac{ \|v_{k}\| }{ \|v_{k}\|_{2} } \to 0 \quad \text{as } k \to \infty. $$
Upon normalizing, we may assume that $\|v_{k}\|_{2} = 1$. Then $\|v_{k}\| \to 0$ and thus $v_{k} \to 0$ w.r.t. $\| \cdot \|$.
On the other hand, by the compactness of $B$, we may assume further (by passing to a convergent subsequence) that that $v_{k} \to v \in B$ w.r.t. $\| \cdot \|_{2}$ This shows
$$ \| v_{k} - v \| \leq C \| v_{k} - v \|_{2} \to 0. $$
This implies $v_{k} \to v \neq 0$ w.r.t. $\| \cdot \|$, a contradiction! Therefore $\text{(*)}$ is proved.
A: How about this: since $DimW=n$ , then $W$ is homeomorphic to $ \mathbb R^n$ , which is complete. This homeomorphism extends into an equivalence/isomorphism as normed spaces, because all norms on a finite-dimensional space are equivalent, see, e.g.:http://planetmath.org/allnormsonfinitedimensionalvectorspacesareequivalent
  Since the two norms are equivalent, the subspace $W$ is also complete, and  complete subspaces are closed. As mentioned above, the two norms are equivalent, so that $W$ is complete and so it is closed. 
