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.
