(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.
(
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.
You will have as many basic variables as constraints, i.e., $3$ here. Typically, a basic variable has a positive value.
If you consider the point $(0,0)$, it means that $x_1=x_2=0$, and that the slack variables $e_1,e_2,e_3$ are positive (for the constraints to hold). So the basic variables are $e_1,e_2,e_3$.
For point $(0,2)$, $x_1=0$ and $x_2>0$, so $x_2$ is a basic variables. You need two more. You can either find them algebraically by plugging the values of $x_1$ and $x_2$ in the constraints: you will find that $e_3=0$, and that $e_1,e_2>0$, so $e_1,e_2$ are your remaining basic variables. You can also work graphically. $(0,2)$ is at the intersection between $x_2=2$ and $x_1=0$, in other words at this point only the third constraint is active, which means that first and second constraints are inactive, i.e., $e_1,e_2 >0$.
Can you do the same for the other points?
Best Answer
No, the solution is not necessarily unique. As an example: $$ \begin{array}{cl} \max_{x,y,z\in\mathbb{R}} & x + y \\ s.t. & \left\{ \begin{array}{l} 0 \leq x \leq 1 \\ 0 \leq y \leq 1 \\ 0 \leq z \leq 1 \\ \end{array} \right. \end{array} $$ which consists in maximizing the sum of the two first coordinate over the simplex. The solutions are of the form $(1,1,z)$, $z\in(0,1)$.
Edit : I am not a specialist of flow network -- at all. Formally, $x$, $y$ and $z$ already enters in two constraints already.
Now, the example was constructed so than an edge of the hypercube is the set of solution (Borisd is right, this edge is orthogonal to the vector defining the objective function). First, you still can deform a little so than this is no longer an hypercube. Secondly, you can also write the same problem in another system of coordinates so that $z$ enters in all the constraints.
More formally, we have an example $$ \min_{x\in\mathbb{R}^n} \ c^Tx \ : \ Ax \leq b $$ which has several solutions. Let $X$ be an invertible matrix. Then the above problem is equivalent to $$ \min_{x\in\mathbb{R}^n} \ c^TXX^{-1}x \ : \ AXX^{-1}x \leq b \ . $$ Now let us consider the problem $$ \min_{y\in\mathbb{R}^n} \ c^TX y \ : \ AXy \leq b \ . $$ This problem has also several solutions. I think that you can choose $X$ so that every component of $y$ enters in all the constraints.