The answer is "more or less yes", depending on your definition of "easily". The following is a possible approach. We use the language of divisors, and we assume the standard facts that the $H^0$ of a skyscreaper sheaf is finite-dimensional and that its $H^1$ is zero.
Assume first that $\mathcal{O}(D)$ is effective. From the short exact sequence
$0 \to \mathcal{O} \to \mathcal{O}(D) \to \mathcal{O}_D \to 0$
and from the vanishing of $H^1(\mathcal{O}_D)$ one obtains a surjective homomorphism
$H^1(\mathcal{O}) \twoheadrightarrow H^1(\mathcal{O}(D))$,
hence the finite-dimensionality of the first group implies the finite-dimensionality of the second.
Now assume that $D$ is any divisor, and write $D=E -F$, with $E$, $F$ effective.
Then by
$0 \to \mathcal{O}(D) \to \mathcal{O}(E) \to \mathcal{O}_F \to 0$
we deduce as before
$H^0(\mathcal{O}_F) \to H^1(\mathcal{O}(D)) \to H^1(\mathcal{O}(E)) \to 0$.
Since $H^0(\mathcal{O}_F)$ and $H^1(\mathcal{O}(E))$ are both finite-dimensional, the claim follows.
EDIT. This argument proves the finite-dimensionality of the sheaf of holomorphic sections. The finite-dimensionality of the sheaf of meromorphic sections $\mathcal{M}(D)$ follows from the fact that the cokernel of the natural injection $\mathcal{O}(D) \to \mathcal{M}(D)$ has trivial $H^1$, see Johannes Ebert's answer.