I'm self-studying Linear Algebra using Axler's Linear Algebra Done Right, and I'm not entirely sure the following proof is correct. I've already seen another solution on StackExchange ( Suppose $V$ is finite-dimensional, Will $L(V,W)$ be infinite-dimensional?) that is clearly cleaner, but I wanted to know how (if at all) the approach I took fails.
Suppose $V$ is finite-dimensional with $\dim V > 0$, and suppose $W$ is infinite-dimensional. Prove that $\mathcal{L}(V, W)$ is infinite-dimensional.
Proof: Let $\dim V = n > 0$, and let $\mathbf{B} = \{v_1, v_2, \ldots, v_n\}$ be a basis of $V$. Now assume for the sake of contradiction that $\mathcal{L}(V, W)$ is finite dimensional. Then we can produce a basis for $\mathcal{L}(V, W)$:$\{T_1, T_2, \ldots, T_m\}$, where $T_i \in \mathcal{L}(V, W) \, \forall i \in \{ 1, \ldots, m \}$. Now consider $\\S = span(T_1 v_1, \ldots, T_1 v_n, \ldots, T_m v_1, \ldots, T_m v_n)$. $S$ is the set of all linear combinations of a finite set of vectors, so it must be finite dimensional. Then we can produce a basis $\{w_1, \ldots, w_k\}$ of $S$, where each $w_i \in W$. But $W$ is infinite-dimensional, so we know there must exist a $w_{k+1} \in W, w_{k+1} \notin span(w_1, \ldots, w_k)$. Now consider $T_{m+1} \in \mathcal{L}(V, W)$ defined by $$T_{m+1}v = \begin{cases}
w_{k+1} &\text{if } v = v_1\\
0 &\text{otherwise}
\end{cases}$$
Note that any linear combination of $\{T_1, T_2, \ldots, T_m\}$ must map to a vector in $S$. However, the image of $T_{m+1}$ includes $w_{k+1} \notin span(w_1, \ldots, w_k) = S$, so $T_{m+1} \notin span(T_1, T_2, \ldots, T_m)$. But this contradicts our assumption that $\{T_1, T_2, \ldots, T_m\}$ is a basis for $\mathcal{L}(V, W)$. It follows that $\mathcal{L}(V, W)$ is infinite dimensional.
Best Answer
I think your proof is generally fine.
You should fix a typo when you define the $T_i$ where you say $i\in\{1,\dots,n\}$. It should be $i\in\{1,\dots,m\}$.
You should make it clear that your definition of the linear transformation $T_{m+1}$ is with regards to the base $B$.
Moreover, you don't need to introduce a basis of $S$, since $S\subset W$ is the span of a finite set and hence must be finite-dimensional. The rest of your argument works perfectly well just from the fact that $\text{Im}(T_{m+1})\not\subset S$.