Homogeneous ideals are contained in homogeneous prime ideals

A different proof: $I$ is contained in a prime (maximal) ideal of $S$, say $P$. Since $I$ is homogeneous, $I\subseteq P^*$, where $P^*$ is the ideal generated by the homogeneous elements of $P$. Note that $P^*$ is also a prime ideal (see Bruns and Herzog, Lemma 1.5.6(a)) and homogeneous.

Let $T$ be the set of all proper homogeneous ideals which contain $I$. Then, by Zorn's lemma, there exists a maximal element of $T$, say $P$. We claim that $P$ is prime. Suppose that for homogeneous elements $a,b \in S$, $ab \in P$ but, $a\notin P$. Then $\langle a \rangle + P$ is a homogeneous ideal which strictly contains $P$. It follows that $\langle a \rangle + P=S$, so $1=ax+p$ with $p\in P$. We thus get $b=abx+bp\in P$. Now use Showing that a homogenous ideal is prime. to conclude that $P$ is prime.