Why would we a priori expect $V(I)$ to satisfy axioms to define the closed sets for a topology on $\text{Proj}(S)$? The topological space $\text{Proj}(S)$ has the underlying set$$\text{Proj}(S) = \{\mathfrak{p} \text{ a homogeneous prime such that }S_+ \not\subseteq \mathfrak{p}\},$$and the closed sets are the loci $V(I) = \{\mathfrak{p} \in \text{Proj}(S) : I \subseteq \mathfrak{p}\}$ for homogeneous ideals $I$ of $S$, where $S$ is an $\mathbf{N}$-graded ring. Perhaps this question is a bit silly since its such a straightforward check, but why would we a priori, intuitively and/or geometrically, expect such $V(I)$'s to satisfy the axioms to define the closed sets for a topology on $\text{Proj}(S)$?
Edit: The comment by Hoot mentions schemes. Could I have an explanation that tries to avoid schemes if possible?
Edit: Someone typed out a perfectly good response only to have deleted it? Why?
 A: The origin of this definition is arguably projective varieties over a, say, algebraically closed field $k$: These are sets of the form $V(I)$ for $I\subseteq k[x_0,\ldots,x_n]$ a homogeneous ideal and
$$V(I)=\left\{ [p_0:\ldots:p_n]\in\mathbb P^n_k \mid \forall d~\forall f\in I_d\colon f(p_0,\ldots,p_n)=0 \right\}.$$
The observation (an implication of the projective Nullstellensatz) is that there is a one-to-one correspondence between points $p=[p_0:\ldots:p_n]\in V(I)$ and maximal relevant homogeneous ideals, where $p$ corresponds to the ideal
$$M_p = \langle  x_ip_j - x_jp_i \mid 0\le i,j\le n  \rangle.$$
Note that $V(M_p)=\{p\}$. With this correspondence, we furthermore have $p\in V(I)$ if and only if $I\subseteq M_p$. Thus, we can give the set of maximal homogeneous ideals of $k[x_0,\ldots,x_n]$ the Zariski topology on $\mathbb P_k^n$ and the closed sets are precisely sets of the form 
$$\mathcal V(I)=\{ M\subseteq k[x_0,\ldots,x_n]\text{ homog. max. relevant ideal} \mid I\subseteq M \}$$
The step from only maximal ideals to prime ideals is an abstraction (and indeed a simplification) that is often helpful in algebraic geometry.
Edit 1: I am rattled by the vote controversy. I myself was not really sure if this answers the OP's question. Please, guys, provide some comments. 
Edit 2: Since the bounty says "Looking for an answer drawing from credible and/or official sources", I suggest the end of "What is algebraic geometry?" in Hartshorne's book. The paragraph starting at the bottom of page 58 has some relation to what I said. 
Alternatively, the beginning of II.2 Schemes is in this spirit:

"[...] to any ring $A$ (recall out conventions about rings made in the 
  Introduction!) we associate a topological space together with a sheaf of 
  rings on it, called $\operatorname{Spec} A$. This construction parallels the construction of affine varieties (I, §1) except that the points of $\operatorname{Spec} A$ correspond to all prime ideals of A, not just the maximal ideals."

A: Like Jesko says, this definition starts with polynomials over an algebraically closed field $k$. One defines the Zariski topology on $k^n$ to have closed sets $\{ (z_1, \ldots, z_n) : f_1(z) = f_2(z) = \cdots = f_r(z) = 0 \}$ for some list of polynomials $f_1$, ..., $f_r$. Why should this be the definition? We want polynomials to be continuous maps $k^n \to k$, and we want $\{ 0 \}$ to be closed in $k$, so these sets have to be closed. If $k$ has no additional structure, there is nothing else which we would obviously want to be closed, and this is already enough to give a topology.
Now, if $f_1(z) = f_2(z) = \cdots = f_r(z) = 0$, then $\sum f_i(z) g_i(z)=0$ for any polynomials $g_i$, so this closed set only depends on the ideal generated by the $f_i$. So we have assigned a closed subset $Z(I)$ in $k^n$ to every ideal $I \subseteq k[x_1, \ldots, x_n]$. We have $Z(I) = Z(J)$ if and only if $\sqrt{I} = \sqrt{J}$ (this is the Nullstellansatz).
How can we intrinsically describe $k^n$, just given $k[x_1,\ldots, x_n]$ as an abstract ring? It turns out (another form of the Nullstellansatz), that every maximal ideal of $k[x_1,\ldots, x_n]$ is of the form $\langle x_1-a_1, \ldots, x_n - a_n \rangle$ for a unique $(a_1, \ldots, a_n) \in k^n$. So, to an arbitrary ring $A$, we can associate the set $\mathrm{MaxSpec}(A)$ of maximal ideals of $A$. For every ideal $I \subset A$, we can define a closed subset $Z_{\max}(I)$ of the maximal ideals by $Z_{\max}(I) = \{ \mathfrak{m} : \mathfrak{m} \supseteq I \}$, and this recreates the given definition for $k[x_1,\ldots, x_n]$.
But, for a general ring, we do not have that $Z_{\max}(I) = Z_{\max}(Z)$ implies $\sqrt{I} = \sqrt{J}$. For example, let $A$ be the localization of $k[t]$ at the prime ideal $(t)$. Then $\mathrm{MaxSpec}(A)$ has only one point (the maximal ideal $(t)$), and both $Z_{\max}((0))$ and $Z_{\max}((t))$ contain that point. Using prime ideals rather than maximal ideals fixes this issue.
Then, finally, we want to go to the homogenous case in order to work with projective space rather than affine space.
