Let $f \colon R \to S$ be an surjective ring homomorphism (for general rings $R$ and $S$, not necessarily integral domains). We want to show that $\dim S \leq \dim R$.
My attempt was as following: The surjective ring homomorphism $f$ induces an injection on the spectra, i.e. an injective map $$\operatorname{Spec} S \to \operatorname{Spec} R.$$ Then we take an arbitrary chain of prime ideals in $S$ and their preimages, which are again prime ideals in $R$. Then since the map on the spectra is injective, no two prime ideals the chain in $S$ can map to the same prime ideal in $R$, so the chain of prime ideals in $R$ is at least as long as the chain in $S$ and hence $\dim S \leq \dim R$. Is this correct or are there any problems I haven't noticed/gaps to fill?
Since $f$ is surjective, $S\cong R/\ker (f)$. Clearly, $\dim (R/\ker (f))\leq \dim (R)$ and so $\dim (S) \leq \dim (R) $.