The exercise is defined as it follows:
Let $G$ connected and $W \subseteq V(G)$ a cut set with minimum cardinality. Show that if $x \in W$ then $\omega(G-W) \leq \delta(x)$.
Where $\omega$ is the number of connected components in $G$.
Proof: Since $W$ is a cut set with minimum cardinality we've got that $|W| = \kappa(G)$ thus, each cut point is contained in at most one connected component, therefore we have: $$ \omega(G - W) \leq \kappa(G) \\ \kappa(G) \leq \delta(G) \\ \delta(G) \leq \delta(x) \\ \therefore \omega(G - W) \leq \delta(x) $$.
I'm not sure about the first argument, where I argue that each cut point is in at most one connected component, also I don't see where I use the hypothesis that if $ x \in W$, why should I use it? Is the proof wrong? If it is, there are any hints?
As a cut set $W \subseteq V(G)$ I shall understand such a set that a graph $G-W$ is not connected. Let $W \subseteq V(G)$ be a cut set with minimum cardinality. Let $G_1,\dots, G_k$ be connected components of the graph $G-W$ and $x\in W$ be any vertex. Since a graph $G_x=G-(W\cup \{x\})$ is connected, for each $G_i$ there is a path $P_i=(v_{i,1}, v_{i,2}\dots, v_{i,l_i})$ of minimal length connecting $x$ and $G_i$, that is $v_{i,1}=x$ and $v_{i,l_i}$ is a vertex of $G_i$. If the degree $\delta(x)$ of the vertex $x$ is less than $k$ then there exists distinct $G_i$ and $G_j$ such that $v_{i,2}= v_{j,2}$. Then a path $(v_{i,l_i},\dots v_{i,2}= v_{j,2},\dots, v_{i,l_i})$ connects $G_i$ and $G_j$ in $G_x-\{x\}=G-W$, a contradiction.