The answer here mentioned a table from Sierksma's $\textit{Linear and Integer Programming: Theory and Practice}$, Volume 1, page 144. Both primal and dual are under standard form in table below (Here is the link to the book):
Primal Optimal Solution Dual Optimal Solution
(a) Multiple implies Degenerate
(b) Unique and nondegenerate implies Unique and nondegenerate
(c) Multiple and nondegenerate implies Unique and degenerate
(d) Unique and degenerate implies Multiple
However, my friend shows me an example seems to conflict with case(d). Here is the primal $$ \begin{aligned} \min \quad x_1 &+ x_2\\ \text{s.t.}\quad \quad & \\ x_1 &= 1\\ x_1 + x_2 &= 1\\ x_1 \geq 0,\ &x_2 \text{ is free.} \end{aligned} $$ and here is the dual $$ \begin{aligned} \max \quad y_1 &+ y_2\\ \text{s.t.}\quad \quad & \\ y_1 + y_2 &\leq 1\\ y_2 &= 1\\ y_1,y_2 &\text{ are free.} \end{aligned} $$
Primal problem has unique degenerated optimal solution $x_1 = 1,x_2 =0$ and the dual also has unique degenerated optimal solution $y_1 = 0,y_2 =1$. (Although with the slack variable, we have two bases corresponding to the dual optimal solution). It seems like there is no other assumptions for the case(d), is there anything wrong with the example above?