I am thinking through a specific case to try and fix some ideas.
Suppose that $ S = \frac{k[x,y,z]}{(x^{2}+y^{2}-z^{2})}, $ and $ S $ is graded in the natural way. One point of $ \text{Proj}(S) $ which is not closed is the generic point $ (0) $ because $ \overline{(0)} = \text{Proj}(S). $ Are there any other non-closed points?
I acknowledge that the closed points of $ \text{Proj}(S) $ correspond to points of $ V(x^{2}+y^{2}-z^{2}), $ and in spite of knowing that maximal ideals classically correspond to points, I feel that scheme-theoretically, I don't truly see why.
As to the homogeneous distinguished open sets:
$ D_{+}(z) = \text{Spec}\Big[S \Big[\frac{1}{z}\Big]\Big]_{0} = \text{Spec}\frac{k[\frac{x}{z},\frac{y}{z}]}{\Big(\frac{x^2}{z^2} + \frac{y^2}{z^2} -1 \Big)}. $
So the $ D_{+}(f) $ is a conic for $ f \in S_{+}, $ and in particular, when $ \text{deg}(f) = 1. $
Note that the zero ideal of $S$ in not prime if $\operatorname{char}k=2$, as then $$x^2+y^2-z^2=(x+y+z)^2.$$ If $\operatorname{char}k\neq2$ then it is prime, which I won't verify here (it isn't hard to do so though).
A non-closed point of $\operatorname{Proj}(S)$ is a non-maximal homogeneous prime ideal of $S$ that doesn't contain the irrelevant ideal $(x,y,z)\subset S$. Because $S/(x,y,z)\cong k$ is a field, there are no non-maximal prime ideals containing the irrelevant ideal. So you are looking for a homogeneous prime ideal of $S$ that isn't maximal.
One example is the ideal $I=(x,y-z)$, which contains $(x^2+y^2-z^2)$ because $$x^2+y^2-z^2=x\cdot x+(y+z)\cdot(y-z).$$ It is a prime ideal because $$S/I=(k[x,y,z]/(x^2+y^2-z^2))/(x,y-z)\cong k[x,y,z]/(x,y-z)\cong k[t],$$ is a domain, and $I$ is non-maximal because $k[t]$ is not a field.
Can you find more examples? And something other than the now obvious $I'=(y,x-z)$?