Let $I\subset J \subset R$ be ideals of $R$. How can we show that $\dim(R/J) \leq \dim(R/I)$?
So far I've shown that the heights meet $ht(I)\leq ht(J)$, which is fairly straight forward, but I am having trouble relating this to chains of prime ideals of the quotient rings.
You correctly deduced the inequality $(1): \, \operatorname{ht}(I) \le \operatorname{ht}(J)$ from the inclusions $(0): \,I \subset J \subset R$. But then you "forget" about $(0)$ and try to deduce $(2): \, \dim R/J \le \dim R/I$ just from $(1)$. But you can not do this in general: there exist instances where (2) does not hold even though (1) hold. That is why you are having trouble relating (1) to (2). Instead, as Krish points out in his comment, just observe that if a prime ideal contains $J$, then it certainly contains $I$. Then apply that to chain of prime ideals containing $J$.