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Every linear function on a finite-dimensional space is continuous.

I was wondering what the domain and codomain of such linear function are?
Are they any two topological vector spaces (not necessarily the same), as along as the domain is finite-dimensional? Can the codomain be a different normed space (and may not be finite-dimensional)?
I asked this because I saw elsewhere the same statement except the domain is a finite-dimensional normed space, and am also not sure if the codomain can be a different normed space (and may not be finite-dimensional).
Thanks and regards!

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 $1\leqslant j\leqslant n$.
