Every quasi-compact scheme has a closed point I know this question has been asked here before, but I have trouble understanding the following proof, taken from a Schwede's write-up. 

I have underlined the bit I don't understand. In particular, I have no idea how we can conclude that $P_3$ is not in $U_2$. Didn't we pick $P_2$ to be any point in the closure of $P_1$? So, can someone explain rigorously why the point $P_3$, which is any point in the closure of $P_2$ (other than $P_2$, right?) ought not to be in $U_2$? 
I know I am missing something obvious here, because this seems to be a common proof. Also, its probably topology that's giving me trouble here, so I will also tag it as such.
 A: Since $P_1$ is a closed point of $U_1$, we have $\overline{P_1} \cap U_1 = P_1$, where $\overline{P_1}$ denotes the closure of $\{P_1\}$ in $X$.
Since $P_2 \in \overline{P_1}$, which is a closed set, we have $\overline{P_2} \subseteq \overline{P_1}$.
Now, I'm not sure if this is strictly necessary, but maybe we're worried that, unlike $P_1$, $P_2$ might not be a closed point of $U_2$.  If not, though, we can replace it with one, because $\overline{P_2} \cap U_2$ is closed in $U$, and a closed set of an affine scheme contains a closed point.  Moreover, since each  point of $\overline{P_2} \cap U_2$ lies in $\overline{P_1}$, we could have chosen this point to being with.
Presumably we're not supposed to take $P_3$ to be $P_1$ or $P_2$, even though it doesn't technically say this.  (Maybe it's automatic that $P_1 \not \in \overline{P_2}$ for some reason I don't see immediately -- I'm having trouble picturing an example.)  Since $P_3 \in \overline{P_2}$, $P_3 \neq P_2$, and $\overline{P_2} \cap U_2 = P_2$, we have $P_3 \not \in U_2$, and similarly $P_3 \not \in U_1$.
