Let $G$ be a graph with at least two vertices throughout this exercise.
(a) Show that $\kappa(G) = 0 \iff \kappa'(G) = 0$.
(b) For any integers $t,s,m$ with $0 < t \le s \le m$, let $G$ be the graph constructed as follows. Take a disjoint union $K \sqcup K'$ of two copies of $K_{m+1}$ (which we will call the left and right cliques respectively). Let $X = \{x_1, \dots, x_s\}$ be $s$ distinct vertices in the left clique and let $Y = \{y_1, \dots, y_t\}$ be $t$ distinct vertices in the right clique. Let $f \colon X \to Y$ be any surjective/onto function (which is possible as $s \ge t > 0$.) Finally, form $G$ from $K \sqcup K'$ by adding a single edge between $x_j$ and $f(x_j) \in Y$ for all $1 \le j \le s$. Show that $G$ is a simple, connected graph which is not complete. Compute its minimum degree $\delta(G)$.
Let $m=s=t=1$, then each X and Y have only one vertex, and by forming a function $f$ between these two vertices, isn't it going to be just complete $K_{2}$ graph?
The graph $G$ is constructed by beginning with two copies of $K_{m+1}$, not of $K_m$. Since $|X|,|Y| \le m$, there will always be at least one vertex in the left clique that's not in $X$, and at least one vertex in the right clique that's not in $Y$.
So you are right that $K_2$ would be a counterexample, but the language of the problem specifically excludes that case.