We could prove $(b)$ and $(c)$ by your inequality chain.
$(b)$ If $G$ is a connected graph and $\Delta$ is an eigenvalue of $G$, then $G$ is regular.
Proof: Let $\lambda=\Delta$ be an eigenvalue of $A$ with eigenvector $v \neq 0$. Choose $j$ such that $| v_j |$ is maximal among all $| v_i |$.
Evaluating $Av = \lambda v$ at $j$ we get $\sum\limits_{i}a_{ij}v_i=\lambda v_j$ and thus
\begin{equation}
\Delta=\lvert\lambda\rvert = \frac{\Big\lvert\sum\limits_{i}a_{ij}v_i\Big\rvert}{| v_j |} \leq \frac{ \sum\limits_{i}a_{ij}|v_i|}{| v_j |} \leq \frac{ \sum\limits_{i}a_{ij}|v_j |}{| v_j |} = \sum_{i}a_{ij} = d(x_j) \leq \Delta\,.\tag{1}\label{eq:main}
\end{equation}
The inequation \eqref{eq:main} becomes an equation exactly when $v_i$'s, whose corresponding $a_{ij}\ne 0$, have same sign due to the first inequality, and $\lvert v_i\rvert$'s must equal $\lvert v_j\rvert$ due to the second inequality. Furthermore, such $v_i$'s and $v_j$ have same sign since $\lambda=\Delta\ge 0$. As a result, $v_i$'s equal $v_j$.
We iteratively continue to consider another row, say, $i$th of $A$, where $v_i$ was determined before, with the same argument. On $i$th row, all eigenvector entries $v_k$'s where $a_{ik}\ne0$ equal $v_i$. Because $G$ is connected, we will entirely traverse all rows. (We can think of this step as Depth-first search or Bread-first search algorithms starting at the $j$th vertex.)
After all, we know that $(v_j,v_j,\ldots,v_j)$ is an eigenvector corresponding to the eigenvalue $\lambda=\Delta$, so does $(1,1,\ldots,1)$. Thus, the sum of each row of $A$ is $\Delta$, which means $G$ is regular.
We similarly prove $(c)$ (with a little bit complexity). We use the same notations in the previous proof, except that $\lambda=-\Delta$.
$(c)$ If $G$ is a connected graph and $-\Delta$ is an eigenvalue of $G$, then $G$ is regular and bipartite.
Proof: For the sake of equality of \eqref{eq:main}, $v_i$'s, whose corresponding $a_{ij}\ne 0$, have same sign due to the first inequality, and $\lvert v_i\rvert$'s must equal $\lvert v_j\rvert$ due to the second inequality. Furthermore, such $v_i$'s have opposite sign with $v_j$ since $\lambda=-\Delta\le0$. Thus, $v_i$'s equal $-v_j$. That implies $a_{jj}=0$ as well.
We iteratively continue to consider another row, say, $i$th of $A$, where $v_i$ was determined before, with the same argument. On $i$th row, all eigenvector entries $v_k$'s where $a_{ik}\ne0$ equal $-v_i$, and $a_{ii}=0$. Because $G$ is connected, we will entirely traverse all rows.
After all, getting an eigenvector whose entries are $v_j$ or $-v_j$, we scale it to one whose entries are $1$ or $-1$. If $v_i=v_k$, $i$th and $k$th vertices are not adjacent, so we partition vertices into 2 sides: one for $v_i=1$, one for $v_i=-1$. In addition, we can easily conclude that the sum of each row is $\Delta$. Consequently, G is bipartite and regular.
