Let $T: V \to W$ be a linear map. Let $V$ be finite dimensional. Then $T$ is continuous. Prove the following, without using that any two norms on a finite dimensional vector space are equivalent.

Let $T: V \to W$ be a linear map between normed spaces. Let $V$ be finite dimensional. Then $T$ is continuous.

My attempt:
I can prove that $T$ is continuous iff $T$ is continuous in $0$. 
So, it suffices to show:
$$\forall \epsilon > 0: \exists\delta > 0: \forall v \in V: \Vert v \Vert <\delta \implies \Vert Tv\Vert < \epsilon$$
Let $E := \{e_i\}_i^n$ be a basis of $V$. Write $v = \sum v_i e_i$
Let $\epsilon > 0$. Choose $\delta := ???$
If $v \in V$ satisfies $\Vert v\Vert < \delta$, then $\Vert Tv\Vert = \Vert \sum v_iT(e_i) \Vert \leq \sum |v_i| \Vert Te_i\Vert< M\sum |v_i|$
where $M := \max_i^n\Vert Te_i \Vert$
However, I don't know how to handle the $\sum |v_i|$. How can I make this small, given that $\Vert v \Vert < \delta?$
 A: My first post here, feel free to correct me:
Suppose $T: V \rightarrow W$ is not continuous and $V$ is finite dimensional. Then there is a sequence $(x_n)_n$ with $||x_n||=1$ and $||Tx_n||\geq n$ for all $n\in\mathbb{N}$. Let be $(e_i)_{i=1}^N$ a basis of $V$. Bolzano-Weierstraß implies the existence of a subsequence $(x_{n_k})_k=(\sum\limits_{i=1}^N a_{i,n_k} e_i)$ converging towards $x=\sum\limits_{i=1}^N a_i e_i \in B_1(0):=\lbrace y \in V \; | \;||y||\leq 1\rbrace$ (which is compact in $V$ iff $V$ is finite dimensional) and with 
$$|a_{i,n_k}-a_i| \rightarrow 0 $$
for all $i=1,...,N$. But we also have
$$0\leq ||Tx||\leq \sum\limits_{k=1}^N |a_i| \cdot||Te_i||\leq C < \infty$$
for some $C\in \mathbb{R}$ and
$$ n_k\leq ||Tx_{n_k}|| \leq |\; ||Tx||-||Tx_{n_k}|| \;|\leq ||Tx-Tx_{n_k}|| \leq \sum\limits_{k=1}^N |a_i-a_{i,n_k}| \cdot||Te_i||$$ 
which implies that $|a_{j,n_k}-a_j|$ can't converge towards $0$ for at least one $j\in \lbrace 1,...,N\rbrace$ because the left side diverges. Contradiction.
Hence $V$ can't be finite dimensional since $B_1(0) \supset \lbrace y \in V \; | \;||y||= 1\rbrace$ can't be compact in $V$.
