= Solution
Here closed oriented surfaces are connected, as in the usual single-genus convention. Let $f:X\to Y$ have nonzero <mapping degree> $d$. Work over $\mathbb Q$, so multiplication by $d$ is invertible and cannot vanish.
The <Poincare duality pairing> on $H^1(Y;\mathbb Q)$ is nondegenerate. Thus for every $0\ne a\in H^1(Y;\mathbb Q)$ there is $b\in H^1(Y;\mathbb Q)$ with $\langle a\smile b,[Y]\rangle\ne0$. Naturality of the <cup product> and evaluation on the <fundamental class> give
$$
\langle f^*a\smile f^*b,[X]\rangle
=\langle f^*(a\smile b),[X]\rangle
=\langle a\smile b,f_*[X]\rangle
=d\langle a\smile b,[Y]\rangle\ne0.
$$
In particular $f^*a\ne0$, so $f^*:H^1(Y;\mathbb Q)\to H^1(X;\mathbb Q)$ is injective. The <cohomology ring of a closed oriented surface> has $\dim H^1(\Sigma_g;\mathbb Q)=2g$. Comparing dimensions now gives
$$
2g(Y)\le2g(X),\qquad\boxed{g(X)\ge g(Y).}
$$
This proves <genus monotonicity under nonzero-degree surface maps> by <rational cohomology injectivity of a nonzero-degree map>, without requiring $f$ to be a covering map or a smooth map.
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