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I think the above answer is enough to prove $ht \; I \le ht \; I' +1 $ if I \neq I'. Pick a prime ideal $q$ that contains $I'$ such that $ht q = ht I'$. Recall that $q$ is homogenous since it is a minimal prime over a homogeneous ideal. By going modulo $q$ we may assume that $q = 0$. Now use the fact stated above by localizing at (0) as above. The image of $I$ is not zero since $I$ in $R_{(0)}$ is not in $q$. And it is not the whole ring since we are only inverting homogeneous elements that are not in $q$ which contains $I'$. Then since $R_{(0)}$ is a PID in particular one diemsimonal domain, the height of the image of $I$ is 1. Now apply the above result again to conclude that $ht I \le ht I' + 1$.