The result we can show is the following:
Let $E$ and $F$ two topological vector spaces, where $E$ is Hausdorff, and $T\colon E\to F$ a linear map. If $E$ is finite dimensional, then $T$ is continuous.
First, if $(e_1,\ldots,e_n)$ is a basis of $E$, then any set of $n+1$ vectors of $T(E)$ is linearly dependent, so $T(E)$ has a dimension $\leqslant n$. Let $k$ be the dimension of $T(E)$, and $(v_1,\ldots,v_k)$ a basis of this space. We can write for any $x\in E$: $T(x)=\sum_{i=1}^ka_i(x)v_i$ and since $v_i$ is a basis each $a_i$ is linear.
We have to show that each map $T_i\colon E\to F$, $T_i(x)=:a_i(x)v_i$ is continuous.
Added: the map $x\mapsto a_i(x)$ is well-defined because $(v_1,\ldots,v_k)$ is a basis. In particular, it takes finite values.
By definition of a topology on a topological vector space we only have to show that the map $x\mapsto a_i(x)$ is continuous. To do that, we use the fact that a finite dimensional Hausdorff topological vector space can be equipped with a norm which gives the same topology (in fact it is the unique one), namely put $$N\left(\sum_{j=1}^n\alpha_j e_j\right):=\sum_{j=1}^n|\alpha_j|$$
Now the continuity is easy to check: denoting $x=\sum_{j=1}^nx_je_j$ and $y=\sum_{j=1}^ny_je_j$
$$|a_i(x)-a_i(y)|\leqslant \sum_{j=1}^n|a_i((x_j-y_j)e_j)|=\sum_{j=1}^n|x_j-y_j|\cdot |a_i(e_j)| \leqslant N(x-y)\sum_{j=1}^n|a_i(e_j)|,$$
since $|x_j-y_j|\leqslant N(x-y)$ for all $1\leqslant j\leqslant n$.
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.
Best Answer
Suppose $X$ infinite dimensional. Then the unit sphere $S$ is not compact (Riesz theorem), and therefore there is a sequence $x = (x_n)_n$ on $S$ without accumulation points. Denote by $x'$ the new sequence defined by $x_n' = (1 + \frac{1}{n})x_n$.
Since $x'$ and $x$ have the same accumulation points, $x'$ doesn't have any. So the set $C$ of the values of $x'$ is closed.
Now $d(0, C) = 1$, and there is no point in $C$ of norm $1$.