Local ring on generic fiber

Let $\pi: X\to C$ be a fibration in curves where $C$ is a non-singular curve and $X$ a regular, integral surface and the generic fiber $X_\eta$ is a non-singular curve over $k(C)$ (these hypotheses might be stronger than necessary, but I just threw a bunch on to make it as nice as possible).

Now a point on $X_\eta$, say $p$ is also a point on $X$ itself. Note the generic point of $X$ is the generic point of $X_\eta$, say $\zeta$. All the other points on $X_\eta$ are closed in the curve and are height $1$ points on $X$.

I read in a paper that for any point on the generic fiber, $p$, we have $\mathcal{O}_{X,p}\simeq \mathcal{O}_{X_\eta, p}$, and at first I just thought to myself that it's obvious, but when I tried to actually think of a reason it wasn't so obvious.

If $X_\eta$ were open in $X$, then this would be clear since restricting to an open and then taking a stalk doesn't cause problems, but why should this still be true for the generic fiber which is neither open nor closed?

One noted consequence is that $k(X)=\mathcal{O}_{X,\zeta}=\mathcal{O}_{X_\eta, \zeta}=k(X_\eta)$.

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Take affine affine open sets $U\subseteq X$ and $V\subseteq C$ such that $\pi|_U: U\rightarrow V$.Then $B:=O_X(U)$ is an $A:=O_C(V)$-algebra and $O_{X,p}=B_p$, where $p\in\mathrm{Spec} (B)$ satisfies $p\cap A=0$.
The generic fibre $X_\zeta\cap U$ equals $\mathrm{Spec}((A\setminus 0)^{-1}B)$. Hence
$O_{X_\zeta ,p} =((A\setminus 0)^{-1}B)_{p(A\setminus 0)^{-1}B}=B_p$
since $B\setminus p$ contains $A\setminus 0$.