Showing Structure Sheaf of Noetherian Scheme Ample Let $X$ consider a noetherian scheme (therefore a quasicoherent one) assump that $H^1(X, \gamma)=0$ holds for every quasicoherent ideal sheaf $ \gamma \subset \mathcal{O}_X$.
The main aim is to show that $X$ is affine but my question refers to following consideration in a reduction step:
Firstly, we show, that the structure sheaf $\mathcal{L} = \mathcal{O}_X$ is, so that for each $x \in X $ there exist a $n \ge 1$ and a $\Gamma(X,\mathcal{L}^{\otimes n}$ such that $x \in X_s$ and $X_s$ is affine.
I would like to know why we can wlog consider only the case that $x \in X$ is a closed point?
 A: I have tried to think about your question. I write it here. Correct me if I am wrong. 
I recall here the definition of Zariski space: A topological space $X$ is a Zariski space if it is Noetherian and every nonempty closed irreducible subset has a unique generic point.
So, in particular, if $X$ is a noetherian scheme then topologically $X$ is a Zariski space (see Exercise 3.17 a), chapter II, page 93, Hartshorne).
Claim: for any point $x \in X$, $\overline{\left \{ x \right \}}$ contains a closed point
For any point $x\in X$, if $x$ is closed, nothing to prove here. Otherwise, consider its closure $\overline{\left \{ x \right \}}$. Suppose that $\overline{\left \{ x \right \}}$ does not contain any closed point, for any point $y$ in $\overline{\left \{ x \right \}}$, consider its closure $\overline{\left \{ y \right \}}$ in $\overline{\left \{ x \right \}}$. Assume that $\overline{\left \{ y \right \}}=\overline{\left \{ x \right \}}$ then by the uniqueness of the generic point, we must have $x=y$, since $y$ is picked randomly we claim that $x$ is closed point which is impossible in this case. So,
$\overline{\left \{ y \right \}}$ is strictly contained in $\overline{\left \{ x \right \}}$. 
Now, if $\overline{\left \{ y \right \}}$ contains a closed point then we are done. Otherwise, repeat the above argument, we get a descending chain of irreducible closed subset of irreducible closed subset of $\overline{\left \{ x \right \}}$. This process must terminate at a finite number of steps by definition of a Noetherian space.
In summary, I have just proved that: for any point $x \in X$, $\overline{\left \{ x \right \}}$ contains a closed point.
For your question: why we only take care of closed points?
For any point $x \in X$, pick a closed point $p_x$ in $\overline{\left \{ x \right \}}$ which is possible by the above discussion. Note that $x$ is the generic point for the irreducible closed subset $\overline{\left \{ x \right \}}$. If you have any open neighborhood $U$ of $p_x$ then $U$ must contains $x$. Why? consider the intersection $V=U \cap \overline{\left \{ x \right \}}$ which is an open set in $\overline{\left \{ x \right \}}$. So, $V$ must be dense in $\overline{\left \{ x \right \}}$ (by a simple contradiction). Hence $x$ must be in $V$ because $x$ is the generic point (by another simple contradiction). Thus, $x \in U$.
In brief, you need only to consider the closed points in $X$.
