Degeneracy in Linear Programming

(Most of this was written before the recent addendum. It addresses the OP's original question, not the addendum.)

(a) Suppose we have distinct bases $B_1$ and $B_2$ that each yield the same basic solution ${\bf x}$. Now, suppose (we're looking for a contradiction) that ${\bf x}$ is nondegenerate; i.e., every one of the $m$ variables in ${\bf x}$ is nonzero. Thus every one of the $m$ variables in $B_1$ is nonzero, and every one of the $m$ variables in $B_2$ is nonzero. Since $B_1$ and $B_2$ are distinct, there is at least one variable in $B_1$ not in $B_2$. But this yields at least $m+1$ nonzero variables in ${\bf x}$, which is a contradiction. Thus ${\bf x}$ must be degenerate.

(b) No. The counterexample linked to by the OP involves the system $$ \begin{align} x_1 + x_2 + x_3 = 1, \\ -x_1 + x_2 + x_3 = 1, \\ x_1, x_2, x_3 \geq 0. \end{align}$$
There are three potential bases in this system: $B_1 = \{x_1, x_2\}$, $B_2 = \{x_1, x_3\}$, $B_3 = \{x_2, x_3\}$. However, $B_3$ can't actually be a basis because the corresponding matrix $\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$ isn't invertible. $B_1$ yields the basic solution $(0,1,0)$, and $B_2$ yields the basic solution $(0,0,1)$. Both of these are degenerate, but there is only one basis corresponding to each.

(c) No. Look at the system $$ \begin{align} x_1 + x_2 = 1, \\ x_2 + x_3 = 1, \\ x_1, x_2, x_3 \geq 0. \end{align} $$ The basic solution $(0,1,0)$ corresponds to bases $\{x_1, x_2\}$ and $\{x_2, x_3\}$. The only other basis is $\{x_1, x_3\}$, which implies that the only other basic solution is $(1,0,1)$. Thus the degenerate basic solution $(0,1,0)$ is not adjacent to another degenerate basic solution.

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(More on part (a), addressing OP's questions in the comments.)

Say there are $n$ total variables in the problem: $x_1, x_2, \ldots, x_n$. Every basis $B$ consists of some $m$ of these variables. The basic solution ${\bf x}$ corresponding to a given basis $B$ has the other $n-m$ variables equal to $0$. (Setting these to $0$ is partly how you determine the value of ${\bf x}$; see, for instance, the examples above). If ${\bf x}$ is degenerate it might have some of the variables in $B$ equal to $0$, too, but the point in terms of the argument is that ${\bf x}$ can have no more than $m$ nonzero variables.

Now, suppose $B_1$ and $B_2$ are distinct and each have $m$ nonzero variables, yet both correspond to ${\bf x}$. Let's say $B_2 = \{x_1, x_2, \ldots, x_m\}$. Since $B_1$ and $B_2$ are distinct, $B_1$ has at least one variable that's not in $B_2$. Let's say this variable is $x_{m+1}$. But since every variable in $B_1$ and $B_2$ is nonzero, that means that $x_1, x_2, \ldots, x_m, x_{m+1}$ are all nonzero. However, $B_1$ and $B_2$ both correspond to ${\bf x}$, which means that there are at least $m+1$ nonzero variables in ${\bf x}$. That cannot happen for a basic solution, and so we have a contradiction.