Proof of rank-nullity via the first isomorphism theorem

I was thinking about the proof of the rank-nullity theorem and I thought about proving it as follows. I just wondered whether this proof worked?

Lemma. If $V$ is a finite-dimensional $F$-vector space and $U\leq V$, then $V/U$ is finite dimensional.

$\hspace{16.5mm}$ Moreover, we have that $\dim{V/U}=\dim{V}-\dim{U}$.

Theorem (Rank-Nullity). If $\alpha:V\to W$ is linear with $V$ finite-dimensional, then
$$\dim{V}=\dim(\text{im }\alpha)+\dim(\ker \alpha)$$

Proof. By the first isomorphism theorem we have
$$V/\ker{\alpha} \cong \text{im }{\alpha}.$$
Taking dimensions and applying the lemma we get
$$\dim V – \dim(\ker\alpha)=\dim(\text{im }\alpha)$$
which on rearrangement yields the result. //

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