You can extract a proof from the classical book "Compact complex surfaces" by Barth-Hulek-Peters-Van de Ven, in chapter VI, section 6 "The Case $a(X)=0$". Here are some details. The proof is quite direct, and does not require quoting the classification of surfaces.

First, one shows that $h^{1,0}(X)\leq 2$ (Proposition 8.1 in Chapter IV of that book).

Assume that $h^{1,0}(X)=0$. From this (and the fact that $K_X\cdot K_X=0$) you deduce immediately from Riemann-Roch that
$H^0(X,-K_X)\neq 0$, which together with $H^0(X,K_X)\neq 0$ implies that $K_X$ is trivial.

The case when $h^{1,0}(X)=1$ cannot happen since it would imply on the one hand that $\chi(\mathcal{O}_X)=1$, while on the other hand $\chi(\mathcal{O}_X)=0$ by the "unbranched covering trick", Proposition 18.1 in Chapter I (plus the fact that $\chi(\mathcal{O}_X)$ behaves multiplicatively under finite unramified covers).

Finally, in the case when $h^{1,0}(X)=2$ one can show that $X$ is a torus by proving that the Albanese map of $X$ is a finite unramified cover. The details are on page 258 of that book
(case e) in their proof).