To find the dual for this linear program, I first represent $y = (y+)-(y-)$ to make sure that every decision variable is greater than or equal to $0$. Then I split the second constraint into \begin{align} 2x + 0.5y &\le 5 \\ -2x - 0.5y &\le -5. \end{align} Then I write out a dual linear program which is a minimizing program with three variables. It's like: $$\min Z = 4a + 5b - 5c$$ Subject to
\begin{align} a + 2b - 2c &\ge -2 \\ a + 0.5b -0.5c &\ge -1 \\ -a - 0.5b + 0.5c &\ge 1 \\ a, b, c &\ge 0 \end{align}
However, this dual program does not have a feasible solution, which should be incorrect. Does anyone have any thoughts? Please help me with this.
Your working seems correct. However, the primal linear program is unbounded, and therefore, by weak duality, the dual is infeasible.
To see that the primal is unbounded, note that for any $y\le 0$, we can set $$x:=(5-0.5y)/2=2.5-0.25y.$$ Then \begin{align} x+y&=2.5-0.25y+y=2.5+0.75y\le 2.5\le 4, \\ 2x+0.5y&=5-0.5y+0.5y=5 \end{align} So $((5-0.5y)/2,y)$ is a feasible solution of the primal for all $y\le 0$. The value of the objective function is $$Z=-2x-y=-5+0.5y-y=-5-0.5y$$ which clearly goes to $+\infty$ as $y\to-\infty$.
Alternatively, you can apply the simplex method to see that the primal is unbounded: you should find a column with negative reduced cost but with no strictly positive entries below it.