In an earlier post Proof of $\mathcal{L}(V,W)$ being infinite dimensional I presented a proof for the proposition that
Theorem. If $V$ is finite dimensional with $\dim V>0$ and $W$ is infinite dimensional then the vector space $\mathcal{L}(V,W)$ is infinite dimensional.
However the proof presented there though correct was preposterously long. I present here another proof which makes use of the following theorem. I would like to know whether this new proof is correct?
Theorem ($2.A.9)$ A vector space $V$ is infinite dimensional if and only if there is a sequence of vectors $v_1,v_2,v_3,v_4,....$ such that for all $n\in\mathbf{Z^+}$ the subsequence $v_1,v_2,....,v_n$ is linearly independent.
Proof. Since $W$ is infinite dimensional we may invoke a sequence of vectors $\psi:\mathbf{Z^+}\to W$ such forall $n\in\mathbf{Z^+}$ the subsequence $\psi(1),\psi(2),...,\psi(n)$ is linearly independent.
Let $v_1,v_2,...,v_m$ be a basis for $V$ and consider the sequence $\phi:\mathbf{Z^+}\to\mathcal{L}(V,W)$ defined as follows $$\phi(n) = T(c_1v_1+c_2v_2+\cdot\cdot\cdot+c_mv_m) = c_1\psi(1)+c_2\psi(2)+\cdot\cdot\cdot+c_m\psi(m+n)\tag{1}$$
We now show by recourse to Mathematical-Induction that with the above definition for all $n\in\mathbf{Z^+}$ the list $\phi(1),\phi(2),...,\phi(n)$ is linearly independent.
Evidently $\phi(1)$ on its own is linearly independent. Now assume that for some $k\in\mathbf{Z^+}$ the list $\phi(1),\phi(2),...,\phi(k)$ is linearly independent but the list $\phi(1),\phi(2),...,\phi(k),\phi(k+1)$ is not.
Then for some $j\in\{1,2,3,...,k+1\}$ it is the case that $\phi(j)\in\operatorname{span}(\phi(1),\phi(2),...,\phi(j-1))$ but this $j\not\in\{1,2,....,k\}$ since $\phi(1),\phi(2),...,\phi(k)$ is linearly independent, consequently $j = k+1$, which implies that for some $x_1,x_2,...,x_k\in\mathbf{F}$ $$\phi(k+1)= x_1\phi(1)+x_2\phi(2)+\cdot\cdot\cdot+x_k\phi(k)$$ then applying both sides to the vector $v_m$ we have $$\phi(k+1)v_m = x_1\phi(1)v_m+x_2\phi(2)v_m+\cdot\cdot\cdot+x_k\phi(k)v_m\tag{2}$$ $$\psi(m+k+1) = x_1\psi(m+1)+x_2\psi(m+2)+\cdot\cdot\cdot+x_m\psi(m+k)\tag{3}$$
but $(3)$ implies that $\psi(m+k+1)\in\operatorname{span}(\psi(1),\psi(2),...,\psi(m+k))$ contradicting the fact that for all $n\in\mathbf{Z^+}$, $\psi(1),\psi(2),...,\psi(n)$ is linearly independent.
$\blacksquare$
I'd do it differently.
Theorem. A vector space $V$ is finite dimensional if and only if no infinite subset of $V$ is linearly independent.
Proof. ($\Rightarrow$) No subset having more than $\dim V$ elements is linearly independent.
($\Leftarrow$) Suppose that $V$ is not finite dimensional. In particular, $V\ne\{0\}$ so we can pick $v_1\in V$, $v_1\ne0$. By assumption, $\langle v_1\rangle\ne V$, so we can pick $v_2\in V\setminus\langle v_1\rangle$ and $\{v_1,v_2\}$ is linearly independent. This starts a recursion: suppose we have picked $\{v_1,\dots,v_k\}$ linearly independent. Since, by assumption, $\langle v_1,\dots,v_k\rangle\ne V$, we can pick a further vector $v_{k+1}\in V\setminus\langle v_1,\dots,v_k\rangle$ so $\{v_1,\dots,v_k,v_{k+1}\}$ is again linearly independent. This procedure doesn't stop, because we cannot find a finite spanning set for $V$. QED
Note that the proof provides, for a non finite dimensional space, an infinite linearly independent subset, which you seem to take for granted.
Now, for your case, since $W$ is not finite dimensional, the procedure above provides an infinite linearly independent subset of $W$, say $\{w_n:n\ge1\}$. Pick a basis $\{v_1,\dots,v_m\}$ of $V$ and define $f_n\colon V\to W$ by $$ f_n(v_1)=w_n, \qquad f_n(v_k)=0\ (2\le k\le m) $$ It shouldn't be difficult to finish up by proving that $\{f_n:n\ge1\}$ is linearly independent in $\mathcal{L}(V,W)$.