Finding Thevenin equivalent circuit and using that circuit when terminals inside circuit
From the circuit drawing, we can see that the "load" is the \$6\Omega\$ resistor. So


  If I do the open circuit way should I omit the 6 ohm resistor and use KCL?


To which the answer is yes. It may not be the easiest way, but it is the most general.

If you have equations where you have more unknowns than equations, then you're probably missing something. Calling the node between the \$15\Omega\$ and \$5\Omega\$ resistor \$v_1\$, the KCL equations in each node would be:

$$\frac{20V-v_1}{15\Omega}=\frac{v_1-v_a}{5\Omega}$$

$$\frac{v_1-v_a}{5\Omega} + 1S\cdot (v_1-v_a) = \frac{v_a}{10\Omega}$$

You also may need to look to your supermesh method. I believe you need to add a contribution in there from your current source. Or

$$V_x = i_1\cdot 5\Omega = I_{sc} - i_1 \Rightarrow I_{sc} = i_1 + i_1\cdot 5\Omega$$