Isn't something like shown below can happen where $d=3$ and $k=1$?
If no, then how do we prove it.
Here v is a vertex with degree $3$, and $C_1$, $C_2$, and $C_3$ are $3$ different components, then removal of $v$ leads to $3$ components.
If $C_1=C_2=C_3=C$ and $C$ is given by the image below.
And $v$ is connected to vertex $u$ of each $C_i$.
Question - Let $d\geq3$ be an integer, and let $G$ be a $d$-regular graph (every vertex has degree $d$) which is $d$-edge-connected.Prove that such a $G$ is tough, meaning that removing any $k$ vertices disconnects $G$ into at most $k$ connected components (for all $k \geq1$).
Let $G$ be a $d$-regular, $d$-edge-connected graph, and suppose that deleting vertices $v_1, v_2, \dots, v_k$ from $G$ leaves connected components $U_1, U_2, \dots, U_m$.
For every $i \in \{1, 2, \dots, m\}$, because $G$ is $d$-edge-connected, there must be at least $d$ edges from $U_i$ to its complement. But since $U_i$ is a connected component of $G - \{v_1, v_2, \dots, v_k\}$, these edges cannot go to any other connected component $U_j$. Therefore there must be at least $d$ edges from $U_i$ to $\{v_1, v_2, \dots, v_k\}$.
Altogether, we get at least $md$ edges between $U_1 \cup U_2 \cup \dots \cup U_m$ and $\{v_1, v_2, \dots, v_k\}$. However, because $G$ is $d$-regular, there can be at most $kd$ edges with an endpoint in $\{v_1, v_2, \dots, v_k\}$. Therefore $md \le kd$, or $m \le k$.
This is exactly the definition of toughness: deleting $k$ vertices from $G$ left at most $k$ connected components.