I don’t see how to justify the statement in question, but I do see how to justify a weaker statement that is still sufficient to prove the theorem.
Suppose that $G-\{x_1,x_2,\ldots,x_k\}$ is connected and and $\deg v_i>k$ for each $i\in\{1,\ldots,k\}$. Then each $v_i$ has at least $k+1$ neighbors in $G$, and at most $k-1$ of them can be in $\{x_1,x_2,\ldots,x_k\}$, so there must be at least two vertices $u_i$ and $v_i$ in $G-\{x_1,x_2,\ldots,x_k\}$ that are neighbors of $x_i$ in $G$. At least one of $u_i$ and $v_i$ must be different from $y_i$; say $u_i\ne y_i$. I claim that this implies that $G-\{x_1y_1,\ldots,x_ky_k\}$ is connected, which is a contradiction.
Let $v$ and $w$ be any two vertices of $G$. If $v,w\notin\{x_1,\ldots,x_k\}$, then there is a path from $v$ to $w$ in $G-\{x_1,\ldots,x_k\}$ and hence in $G-\{x_1y_1,\ldots,x_ky_k\}$. If $v=x_i$ and $w\notin\{x_1,\ldots,x_k\}$, there is a path $P$ from $w$ to $u_i$ in $G-\{x_1y_1,\ldots,x_ky_k\}$, and we can add the edge $u_ix_i$ to $P$ to get a path from $w$ to $x_i$ in $G-\{x_1y_1,\ldots,x_ky_k\}$. Finally, if $v=x_i$ and $w=x_j$ with $i\ne j$, there is a path $P$ from $u_i$ to $u_j$ in $G-\{x_1y_1,\ldots,x_ky_k\}$, and we can extend it with the edges $x_iu_i$ and $u_jx_j$ to get a path from $x_i$ to $x_j$ in $G-\{x_1y_1,\ldots,x_ky_k\}$. This proves the claim.
This contradiction shows that there must be at least one $i\in\{1,\ldots,k\}$ such that $\deg x_i\le k$, and since we already know that $k=\lambda(G)\le\delta(G)\le\deg x_i$, we must have $\deg x_i=k$. Deleting the neighbors of $x_i$ disconnects $G$, so $\kappa(G)\le k=\lambda(G)$.
Would separating $A$ into two nodes connected by one edge work? Also, the minimum degree of your graph is $2$ at each of the four corner nodes.
The graph with the largest minimum degree, lowest edge connectivity, and lowest vertex connectivity is formed by connecting two complete graphs with one singular edge like a bridge.
Best Answer
The inequality $\kappa'(G)\leqslant \delta(G)$ holds for any connected graph, since the edges incident to a vertex of minimum degree form an edge cut. For the reverse inequality, let let $G=(V,E)$ be a graph with $\operatorname{diam}(G)=2$ and $[S,\overline S]$ be a minimum edge cut with $|S|\leqslant |\overline S|$. Then every vertex in $S$ is adjacent to a vertex in $\overline S$. For if not, there would be a vertex $u\in S$ such that for each $v\in\overline S$, there is a path $u\to w\to v$ in $G$, which implies that $$|[S,\overline S]|\geqslant |\overline S|\geqslant \frac{|V|}2. $$ Then $\deg u\leqslant |S|-1<\frac{|V|}2$, so that $$\delta(G)\leqslant d(u)<\frac{|V|}2\leqslant |[S,\overline S]|=\kappa'(G), $$ a contradiction. Since $|[S,\overline S]|= \sum_{v\in S}\deg v - 2e(G[S])$ , it follows that \begin{align} |[S,\overline S]|&\geqslant |S|\delta(G) - |S|(|S|-1)\\&\geqslant |S|\delta(G)-\delta(G)(|S|-1)\\&=\delta(G), \end{align} and hence $$\delta(G)\leqslant |[S,\overline S]|=\kappa'(G). $$