$X$ is a minimal separating set iff every vertex in $X$ has a neighbor in the components of $G-X$ containing $u$ and in the component containing $v$ I'd like to prove the following problem:
Let $G$ be a graph and suppose some two vertices $u,v \in V(G)$ are separated by $X \subseteq V(G)$. Show that $X$ is a minimal separating set $\iff$ every vertex in $X$ has a neighbor in the component of $G-X$ containing $u$ and another in the component containing $v$.
The solution I have proves the problem as follows:
Let $X$ be a separating set of $u$ and $v$. This means that $u$ and $v$ are in different components of $G-X$. Let's call these components $C_u$ and $C_v$. Let $x \in X$. If $x$ is connected to both $C_u$ and $C_v$ in $G$ then adding it to $G-X$ connects the components $C_u$ and $C_v$ i.e. $X-x$ is not separating. If $x$ is not connected to both, say it has no neighbors in $C_v$, then adding it back to $G-X$ cannot joint $C_u$ and $C_v$. Therefore, $X-x$ is still separating.
Now comes the part that confuses me:
This means that removing a vertex $x$ from a separating set keeps the set separating $\iff x$ had neighbors in both $C_u$ and $C_v$. Hence $X$ is minimal separating $\iff$ all its vertices have neighbors in both $C_u$ and $C_v$.
Question: $X$ stays separating if the vertex $x$ is not connected to both. How can I conclude then that removing $x$ keeps the set separating $\iff$ $x$ had neighbors in both $C_u$ and $C_v$.  What is the reasoning behind the final conclusion?

The sentence is just wrong; instead of "$x$ had neighbors" it should be "$x$ did not have neighbors".
