The statement of part (a) of this proposition is as follows:
Let $X$ be a noetherian integral separated scheme which is regular in codimension 1. Let $Z$ be a proper closed subset of $X$, and let $U=X\setminus Z$. Then there is a surjective homomorphism $\mathrm{Cl}(X)\to\mathrm{Cl}(U)$ defined by $\sum n_i Y_i\mapsto\sum n_i(Y_i\cap U)$, ignoring empty intersections.
Now, this seems like it should be blatantly obvious, and the proof in the book more or less treats it as such. I'm having trouble with the first line:
If $Y$ is a prime divisor on $X$, then $Y\cap U$ is either empty or a prime divisor on $U$.
Clearly the former possibility is trivial. For some reason, I'm having trouble with the latter. A prime divisor is defined to be a closed integral subscheme of codimension one. Now, obviously $Y\cap U$ is closed in $U$. But to refer to it as a closed subscheme, don't I need to know what the subscheme structure is? Earlier they refer to the induced reduced closed subscheme structure; I guess I'm supposed to assume this is what they mean, but I'm not 100% sure (though I don't know what else makes sense).
Moving past this for now, I need to know that $Y\cap U$ is integral in $U$ and that $\mathrm{codim}(Y\cap U,U)=1$. Maybe I'm just confusing myself here; since $Y$ is integral (it's a prime divisor on $X$) and $Y\cap U$ is open in $Y$, we know $Y\cap U$ is integral as a subscheme of $Y$; that is, $(Y\cap U, \mathcal{O}_Y|_{Y\cap U})$ is an integral scheme. But is this the same as the closed subscheme structure of $Y\cap U\subseteq U$? Does it even matter?
Finally, the codimension; $\mathrm{codim}(Y,X)=1$ means that there are no irreducible closed subsets of $X$ strictly between $Y$ and $X$ (both are integral, hence irreducible themselves). I think I can see at least why this is true. Assuming we know $Y\cap U$ is an integral (hence irreducible) closed subscheme of $U$, either its codimension is one or it equals $U$. But the latter would force $U\subseteq Y$, which would make $X=Y\cup Z$ not irreducible, which would be a contradiction.
I'd greatly appreciate it if someone could point me in the right direction here; am I way overthinking this? Or are there really some subtleties that I'm not seeing?
Let X be a an integral scheme, Y a closed integral subscheme of codimension 1, and U a nonempty open subscheme. With the topology induced from U, $Y \cap U \subset U$ is closed, and with the reduced subscheme structure induced from U, it becomes a reduced subscheme of U. With the topology induced from Y, $Y \cap U \subset Y$ is irreducible, as a nonempty open subset of an irreducible space. Since Y and U induce the same topologies on $Y \cap U$, one sees that as a scheme $Y \cap U$ is integral (reduced and irreducible).
Now $\mathrm{codim}(Y \cap U, U) = \mathrm{codim}(Y, X) = 1$ follows from the fact that there is a bijection $Z \mapsto \overline{Z}$ from the set of closed irreducible subspaces of U to the set of closed irreducible subspaces of X intersecting U, which in fact maps irreducible components to irreducible components. (See (EGA, 0_IV, 14.2.3).)