Finite-dimensional subspace normed vector space is closed I know that probably this question has already been answered, but I'd like to present my attempt of solution.
Let $(E,\|\cdot\|)$ be a normed vector space and let $F\subseteq E$ be a finite-dimensional subspace. Suppose that $\dim_{\mathbb{R}} F=n$.
First of all I proved the following lemma.
Lemma 1 Every finite-dimensional normed vector space is complete.
Then, I noticed that $(F,\|\cdot\|_{F})$ is a finite-dimensional normed vector space, where $\|\cdot\|_{F}$ is the induced norm. So, $F$ is complete by Lemma 1. 
Lemma 2 If  $(E,\|\cdot\|)$ is a normed vector space such that $\dim_{\mathbb{R}}{E}=n$, then $E$ is algebraically and topologically isomorphic to $\mathbb{R}^{n}$.
By Lemma 2, I can view $F$ as a complete subspace of $\mathbb{R}^n$ and, since $\mathbb{R}^n$ is itself complete, I can conclude that $F$ is closed. Here I used the fact that every complete subspace of a complete space is closed.
Is there something wrong?
 A: The part with lemma 2 isn't necessary. Any complete subspace of a metric space is closed. So the proposition follows directly from lemma 1 and you don't need to assume that $E$ is finite-dimensional. But can you prove lemma 1?
Actually even if you use just “complete subspace of complete space is closed” you don't need lemma 2 since both $E$, $F$ are complete by lemma 1.
On the other hand lemma 1 is a consequence of lemma 2. That's because a homeomorphism between normed linear spaces which is also linear isomorphism preserves completeness. It is so because the mapping is then bilipschitz which goes from the following fact:
For a linear map between normed linear spaces the following are equivalent:


*
  
*it is bounded
  
*it is lipschitz
  
*it is continuous
  
*it is continuous at zero


The only non-trivial implication is $(4) \implies (1)$. By continuity at zero, there is $δ > 0$ such that $f[B(0, δ)] ⊆ B(f(0), 1)$. But then $f[B(0, 1)] ⊆ B(f(0), \frac{1}{δ})$ and $\lVert f(x)\rVert ≤ \frac{1}{δ}\lVert x \rVert$ for every $x$ by linearity.
