When the sheafification map is a presheaf epimorphism Claim: Suppose $\mathcal{F}$ is a glueable presheaf on a paracompact hausdorff space. Then the sheafication map on global sections $\mathcal{F}(X) \to \tilde{\mathcal{F}}(X)$ is surjective. 
(Note this implies that if $X$ is hereditarily paracompact, like for instance if $X \subset \mathbb{R}^n$, then the map is surjective over all open sets $U \subset X$.)
Here is the proof, from Ramanan's book. The idea of the proof is a little tricky, involving a "shrinking". I think a lot of people would chalk it up to weird point-set topology, and not pay too much attention to it. I'm trying to get a better feel for it. Is this indicative of how shrinkings are used in general? What feels mysterious to me is that we're able to disregard somehow the $U_i$'s where $x \in \overline{U}_i \setminus U_i$.
Perhaps someone can give me some useful situations where "shrinking" arguments like this one are used.
A more intuitive proof than the one in the link would also be welcomed!
I put below the fold my current attempt to understand the idea of the proof. (Please know that the mechanics of the proof make sense to me; I just don't "grok" it.)

It seems that we start with some locally finite covering $U_i$, and sections $f_i$ of the presheaf, which agree on stalks. We'd like to use the fact that finitely many sections which share the same germ at $x$ have a common representative on some open set about $x$. Therefore we want to choose a neighborhood $N_x$ for each point that meets only finitely many of the $U_i$. But we have to worry about $x$ being in $\overline{U}_i \setminus U_i$ for some $i$, in which case there is no germ at $x$ corresponding to the index $i$. Therefore we choose the shrinking $V_i$. 
My thought right now is to think of the $\overline{V}_i$ like "indicators": we want to say it's OK if we bump into $U_i$ a little, but not too much, and we choose $N_x$ so that if $N_x$ meets $\overline{V}_i$, then $N_x$ is contained entirely in $U_i$.
 A: This is what I've come up with after some thought.
If we're given an element in the sheafification (a compatible collection of germs represented by $f^i, U_i$ agreeing on germs), we would seek to find another cover with the same germs, which actually agreed as \textit{sections}. The ability to choose a locally finite cover will be key here, but not for the reason that it might seem.
There's a seductive line of argument that says: Take a locally finite subcover of $U_i$, use that all the $f^i$ agree on germs, and use that I can choose a common representative $g^x$ for finitely many sections that have the same germs at $x$. 
This kind of misses the point: all the $f^i$ have the same germs, so getting the $g^x$ to agree with every $U_i$ defined at $x$ is overkill: just agreeing with one will give $g^x$ the right germs. Rather, the place where local finiteness comes in is in choosing the $g^x, V_x$ so that whenever two of them meet, they actually agree with each other. To do this, make it so that whenever $V_x$ and $V_y$ meet, they are actually restrictions of some common $f^i$, where $U_i \supset V_x \cup V_y$. 
Here we are going to use the agreement on the germs: to make each $g^x$ the restriction of a bunch of the $f^i$ simultaneously, so that it can be versatile in what other $g^y$ it agrees with. Specifically, take for each $x$ a finite subset $I_x$ of the index set $I$ of $U_i$, and choose $g^x$ to be a restriction of all the $\{f^i, U_i \mid i \in I_x \}$ simultaneously. The set $I_x$ will need to be such that 


*

*If $i \in I_x$, then $V_x \subset U_i$.

*If $M_x$ meets $M_y$, then $I_x$ meets $I_y$ (so they are both restrictions of some common $f^i, U_i$). 


Pass to a locally finite refinement of the $U_i$, and choose a locally finite closed refinement $Z_i$ of that. (We can also make it so that the interiors of $Z_i$ cover $X$, but we won't need that here.) Then define $I_x = \{i \in I: x \in Z_i\}$. It is a virtue of a locally finite closed cover that we can choose a neighborhood $V_x$ around each $x$ that only meets $\{Z_i: i\in I_x\}$, because $\bigcup\limits_{\substack{\text{ones not} \\ \text{containing} \\ x}}Z_i$ is closed, by local finiteness. 
Now note that $I_x$ satisfies the above requirements, as long as we choose $V_x$ to also be inside $\cap_{i \in I_x} U_i$, which is open since $I_x$ is finite. (Requirement 2 comes because if $z \in V_x \cap V_y$, then just take $i \in I$ such that $z \in Z_i$, and then $V_x \cup V_y \subset U_i$.)
The beauty of the shrinking $Z_i$ is that we can kind of play the $U_i$ and the $Z_i$ off of one another: we can say that all the "tags" $i \in I_x$ we've chosen are such that $V_x \subset U_i$, but at the same time, we've managed to tag every $i$ where $Z_i$ meets $M_x$. (Requirement 1 above wants us to tag fewer, where requirement 2 wants us to tag more.) 
