Understanding why every 3-connected graph that is not $K_4$ has an edge such that suppressing an end also gives a 3-connected graph

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I am reading Diestel graph theory rn.

$\newcommand{\dotdiv}{\mathbin{\rlap{\;{}^{\boldsymbol{\cdot}}}-}}$The definition of $G\dotdiv e$ is as follows: it is the multigraph obtained from $G-e$ by suppressing any end of $e$ that has degree two in $G-e$.

Here is the lemma and the proof that I am trying to understand

Where it says "since G is 3-connected, the vertices subdividing an edge of $J$ could be joined by an $H$-path to a vertex not on the same subdivided edge of $J$", I don't understand why this is true.

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By an $H$-path, we mean a path intersecting $H$ only in its ends. If $H\neq J$ then some edge of $J$ is subdivided to form $H$, by a vertex $u$. Now $G$ is 3-connected, so we can find three paths between any two vertices of $G$. But $u$ has degree 2 in $H$, so some path has interior outside $H$.