It suffices to show that they have the same set of closed sets. We note the closed sets in Zariski topology of Proj S simply as closed sets, and the ones in subspace topology as induced closed sets.
The closed sets are of the form $V_+(I)=V(I)\cap \rm{Proj}\ S$ where $I$ is a homogeneous ideal of $S$.
The induced closed sets are of the form $V(I)\cap \rm{Proj}\ S$ where $I$ is a general ideal of $S$.
Clearly every closed set is also an induced closed set, but I can't show the other way or give a counterexample.
In the end, I tried to show $V(I)\cap {\rm Proj\ S}=V(I^h)\cap {\rm Proj \ S}$.
Yes, they are the same. But not in the sense $V(I)\cap {\rm Proj \ S}=V(I^h) \cap {\rm Proj \ S}$.
Given a general ideal $I$ of $S$, we intend to find a homogeneous ideal $J$ of $S$ s.t. $V(I)\cap {\rm Proj \ S}=V(J) \cap {\rm Proj \ S}=V_+(J)$. There are two ways to produce a homogeneous ideal from a general ideal, and normally only one of them will be mentioned in textbooks.
The famous one is the homogenization of an ideal $I$, denoted $I^h$, defined to be generated by the set of homogeneous elements of $I$.
I don't even know the name of the other one, personally I would call it the outer homogenization of $I$, denoted by $I_h$, defined to be generated by the set $$\{h\ \text{homogeneous} \mid \exists f \in I \text{ s.t. h is one of the homogeneous components of } f\}.$$
Next we show $V(I)\cap {\rm Proj \ S}=V(I_h) \cap {\rm Proj \ S}$ which will complete the proof.
By construction, we have $I\subset I_h$, hence $V(I)\supset V(I_h)$ so $V(I)\cap {\rm Proj \ S}\supset V(I_h) \cap {\rm Proj \ S}$.
Reversely, let $P$ be a homogeneous prime ideal containing I but not containing $S_+$. We intend to show that $P$ contains the generating set of $I_h$, which implies $P\supset I_h$, then the result follows.
Pick a homogeneous element $h$ in the generating set, so there exists $f\in I$ s.t. $h$ is one of the homogeneous components of $f$. Since $f\in I\subset P$, by the homogeneous property of $P$, we must have all homogeneous components of $f$ contained in $P$, in particular we have $h\in P$. The result follows.