The previous post was not written correctly. So, I am posting it again with the correct format I originally wanted.
Exercise: Suppose $V$ is finite-dimensional with $\dim V >0$, and suppose $W$ is
infinite-dimensional. Prove that $L(V, W)$ is infinite-dimensional.
Proof: Let $v_1,\dots,v_n$ be a basis of $V$. Suppose that $L(V, W)$ is finite dimensional. Then there exist linear maps $T_1,\dots,T_k$ such that for all $S\in L(V, W)$ we have that $S=a_1T_1+\dots+a_kT_k$. Using the theorem that there exists a unique linear map that can take on any value on its basis, we can define many such linear maps $S_j$ for $j\in \Bbb{Z^+}$ such that for every $w\in W$ there exists a linear map $S_j$ such that for one fixed vector $v_m$ in the basis of $V$ we have that $S_j(v_m)=w$.
By hypothesis we know that $T_1,\dots,T_k$ spans $L(V, W)$. Thus there exist scalars $a_1,\dots,a_n\in F$ such that $S_j(v_m)=a_1T_1(v_m)+\dots+a_kT_k(v_m)$. Because we have a map $S$ for every $w\in W$ and $T_1(v_m),\dots,T_k(v_m)\in W$. The above representation implies that $T_1(v_m),\dots,T_k(v_m)$ spans $W$. Which is a contradiction as $W$ is infinite dimensional. Hence, $L(V, W)$ is infinite-dimensional.
Is the proof correct?
Edit: Accidentally used the same variable for the length of the basis of $L(V,W)$ that I used for the basis of $V$. Changed the variables now.
Best Answer
I think you have made things more complicated than they should be.
As $W$ is infinite-dimensional, for every $m$ there must exists a linearly independent set in $W$ with $m$ elements, lets say $\{w_1, \cdots, w_m\}$.
Let $\{v_1, \cdots, v_n\}$ be a basis for $V$, and define (using the theorem you mention in your solution) $f_i \in L(V,W)$ for $i=1, \cdots, m$ given by $f_i(v_1)=w_i$ and any other vectors of $W$ for the rest of the $v_i$.
Then, $\{f_1, \cdots, f_{m}\}$ is a linearly independent set in $L(V,W)$ of $m$ elements. As $m$ is arbitrary, $L(V,W)$ cannot have finite dimension.