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Let $\mathcal{O}$ be a Noetherian, one dimensional domain and $P\subset \mathcal{O}$ be a prime ideal. I want to prove that if $0\neq I\subseteq P$ we have $\mathcal{O}/(\mathcal{O}\cap I\mathcal{O}_P) \cong \mathcal{O}_P/ I\mathcal{O}_P$.

It's clearly enough to prove that the map $\mathcal{O}\rightarrow \mathcal{O}_P/ I\mathcal{O}_P$ is surjective, but I can't prove that.

This is used in the proof of propostion 12.3 in Neukirch Algebraic Number Theory.

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    Related: http://math.stackexchange.com/questions/1500673/why-does-mathfrakp-supseteq-mathfraka-imply-mathfrakp-is-the-only – 2017-02-24
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    Would the Noetherian tag be appropriate? – 2017-02-24

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Since $\mathcal{O}$ is a $1$-dimensional domain and $I\neq 0$, $\mathcal{O}/I$ is $0$-dimensional. Since it is Noetherian, that means $\mathcal{O}/I$ is the product of its localizations at each of its prime ideals. In particular, the map from $\mathcal{O}/I$ to any localization of $\mathcal{O}/I$ is surjective (since it is just the projection onto a factor of a product). Thus the map $\mathcal{O}/I\to \mathcal{O}_P/I\mathcal{O}_P$ is surjective, which gives the isomorphism you want.