Is the Zariski Topology of Proj S the same as the subspace topology of Spec S? 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 $V(I)\cap {\rm Proj\ S}=V(I^h)\cap {\rm Proj \ S}$.
 A: 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.
