The total wave function needs to be antisymmetric under particle interchange. Since each electron is in the same 1-particle ground state, $E_0(x)$, the spatial wave function will be symmetric under interchange; hence, the spin wave function must be antisymmetric.
The 2-particle wave function is:
$$ E_{0,0}(x_1, x_2) = E_0(x_1)E_0(x_2) = E_0(x_2)E_0(x_1) = E_0(x_2, x_1).$$
The rules regarding addition of angular momentum are well documented. The antisymmetric ground state will have $S=0$, and of course $S_z=0$:
$$\Xi_{1, 2} = \frac 1{\sqrt{2}}(\uparrow_1\downarrow_2-\downarrow_1\uparrow_2),$$
where the subscripts label the particle index (and the arrows indicate the z-component). Note that:
$$\Xi_{2, 1} = \frac 1{\sqrt{2}}(\uparrow_2\downarrow_1-\downarrow_2\uparrow_1) = -\frac 1{\sqrt{2}}(\uparrow_1\downarrow_2-\downarrow_1\uparrow_2)=-\Xi_{1, 2}$$
so that the spin state is indeed antisymmetric.
The total wave function is their product:
$$ \psi_{1, 2} = E_0(x_1)E_0(x_2)\Xi_{1, 2}.$$
Note that the statement "the electrons have different spin" is misleading (I would even say "classical"): they have the same spin: $J = \hbar\sqrt{j(j+1)} = \sqrt{3/2}\hbar$. They even have the same projection onto the $z-$axis: $\pm\hbar/2$--it's just that their combination is antisymmetric under interchange.
Finally: nowhere did I need to refer to the quantitative solution of the square well.
