In LADR by Axler, there are two theorems stated as following:
Theorem 1: Two finite-dimensional vector spaces over $\mathbb{F}$ are isomorphic $\iff$ They have the same dimension.
Theorem 2: $V$ and $W$ finite dimensional $\implies$ $\mathcal{L}(V,W)$ is finite and $\dim\ (\mathcal{L}(V,W))=(\dim\ > V)(\dim\ W)$
I suppose one way to show isomorphism is by showing a bijection, but in light of the two theorems above should it not be possible to just state that they have the same dimension instead?
As discussed in the comments, the askers proof is valid in the case that $V$ is a finite vector space. unfortunately, if $V$ is infinite one must argue differently. Fortunately, there is canonical bijection between the two spaces $V$ and $\mathcal L(\mathbb F,V)$ that is simple to construct. We perform such a construction in the result below which yields the required result as an immediate corollary:
Result:
For all fields $\mathbb F$ and vector spaces $V$ thereover, the map $\phi:\mathcal L(\mathbb F,V) \to V \ T \mapsto T(1_\mathbb F)$ is an isomorphism of vector spaces.
Proof
Suppose that $\mathbb F$ and $V$ are as given. We first prove that $\phi$ is a linear map. To that end suppose that $T,T^\prime \in \mathcal L(\mathbb F,V)$ and $\lambda, \lambda^\prime \in \mathbb F$. Remark by algebraic manipulation that: $$ \begin{align} \phi(\lambda \cdot T + \lambda^\prime \cdot T^\prime) &= \left(\lambda \cdot T + \lambda^\prime \cdot T^\prime\right)(1_\mathbb F)\\ &= \lambda \cdot T(1_\mathbb F) + \lambda^\prime \cdot T(1_{\mathbb F})\\ &= \lambda \cdot \phi(T) + \lambda^\prime \cdot \phi(T^\prime). \end{align} $$ This immediately verifies that $\phi$ is a linear map. To complete the proof it remains to prove that $\phi$ is a bijection of underlying sets. We do this by showing that $\phi $ is both injective and surjective.
To se that $\phi$ is injective, suppose that $T,T^\prime \in \mathcal L(\mathbb F,V)$ are such that $\phi(T) = \phi(T^\prime)$. It suffices to prove that $T = T^\prime$. In this case suppose that $a \in \mathbb F$. Observe equation $(1)$ follows by the linerity of $T$ and $T^\prime$ and the hypothesis that $\phi(T) = \phi(T^\prime)$:
$$\begin{align} T(a) &= T(a \cdot 1_{\mathbb F})\\ &= a \cdot T(1_\mathbb F)\\ &= a \cdot \phi(T)\\ &= a \cdot \phi(T^\prime)\\ &= a \cdot T^\prime(1_\mathbb F)\\ &= T^\prime(a \cdot 1_\mathbb F)\\ &= T^\prime(a). \tag{1} \end{align}$$ Remark that we have shown the statement $\forall a \in \mathbb F: T(a) = T^\prime(a)$. Thus, by function extentionality, $T = T^\prime$. This confirms that $\phi$ is injective.
We complete the proof by showing that $\phi$ is surjective. Suppose that $a \in \mathbb V$. Let $T: \mathbb F \to V \ x \mapsto x \cdot a$. It is a very quick exercise to see that $T$ is linear. Note now that $\phi(T) = T(1_\mathbb F) = a$. Remark we have shown the statement that $\forall a \in V \ \exists T \in \mathcal L(\mathbb V,\mathbb F): \phi(T) = a$. This is exactly to say that $\phi$ is surjective and so the result is proved.