The proof of Fan lemma in graph theory: a step that I have trouble understanding. So I was reading the following proof of the lemma usually known as the Fan lemma in graph theory:
https://math.stackexchange.com/a/1300354/348091
If $G$ is a $k$-connected graph. $x \in V(G)$ is a vertex, $V \subseteq V(G)$, and the size of $V$ is at least $k$, and furthermore $V$ does not contain $v$. Then there are $k$ vertex-disjoint paths from $x$ to $V$ except at $x$.
The proof in the link uses the following strategy: expand $G$ by adding another vertex $y$ and connecting it to all vertices in $V$; call this graph $G'$ and clearly $G'$ is also k-connected. By Global Menger's Theorem, there are $k$ internally vertex-disjoint paths from $x$ to $y$.
So far the proof makes sense to me. However I've spent hours trying to understand the next step but haven't succeeded at all.
"When we remove $y$ from each of these paths we obtain the $k$ paths that are necessary for the fan lemma."
I can't understand why we have got the desired result by this step: true, we have the $k$ vertex-disjoint paths from $x$ to $y$, but $y$ is not in $V$ and if all those $k$ paths have no internal vertices in $V$, then removing $y$ destroys the paths from $x$ to $V$.
Please help me understand this proof. Thank you very very much.

All of those $k$ internally disjoint paths $P_1,\ldots, P_k$ do have internal vertices in $V$ though, as $y$ is adjacent only to vertices in $V$. Meanwhile $y$ is an **endpoint of each of the $P_i$s so I am not sure what you mean by destroying these paths.
More specifically: Let $P_1,\ldots, P_k$ be the internally vertex-disjoint paths from $x$ to $y$. Then each $P_i$ can be written $xu_1u_2\ldots u_{\ell_i}v_iy$ where $v_i$ is in $V$ [as those are the only vertices adjacent to $v$], and where $i \not = j$ implies $v_i \not = v_j$ [as the $P_i$s are internally vertex-disjoint]. So $P'_1,\ldots, P'_k$ where $P'_i = xu_1u_2\ldots u_{\ell_i}v_i$ are paths that are vertex-disjoint from $x$ to $V$ except at $x$.
