The particle of mass $m$ in the box of length $L$ in 1D is solved by wavefunctions
$$
\begin{align}
\psi_{n\alpha}&=A\sin (k_n x) e^{-\omega_n t}|\alpha \rangle\;, \\
k_n&=\frac{n\pi}{L}\;,\\
E_n&=\hbar \omega_n\;,\\
\omega_n&=\frac{\pi h n^2}{4L^2m}\;.
\end{align}
$$
Here, $|\alpha \rangle$ represents the spin state.
The global fermionic wavefunction for two particles is constructed from all pairs by anti-symmetrization, as
$$
\Psi_{n\alpha m\beta}(x_1,x_2,t)=\psi_{n\alpha}(x_1,t)\psi_{m\beta}(x_2,t) - \psi_{m\beta}(x_1,t)\psi_{n\alpha}(x_2,t)\;.
$$
Energy of state $\Psi_{n\alpha m\beta}(x_1,x_2,t)$ can be calculated as
$$
(H_1+H_2)\Psi_{n\alpha m\beta}(x_1,x_2,t)=(E_n+E_m)\Psi_{n\alpha m\beta}(x_1,x_2,t)\;,
$$
since each of the one-particle Hamiltonians acts on the respective one-particle wavefunction $\psi_{n\alpha}(x_1,t)$, which yields its eigenenergy $E_n$.
For identical spins, we are interested only in solutions for which $\alpha=\uparrow$ and $\beta=\uparrow$. The ground state is the lowest lying energy state of the system. In this case, it would correspond to $\Psi_{1\uparrow 1\uparrow}$, but this function is identically zero. Then next two lowest lying states are $\Psi_{1\uparrow 2\uparrow}$ and $\Psi_{2\uparrow 1\uparrow}$. Thanks to the antisymmetrization, $\Psi_{1\uparrow 2\uparrow} = -\Psi_{2\uparrow 1\uparrow}$ and it represents the ground state of the system with energy $E_1+E_2$. For opposite spins, we choose $\alpha=\uparrow$ and $\beta=\downarrow$. Here, the lowest lying energy state is $\Psi_{1\uparrow 1\downarrow}$ and it has energy $2E_1$.
From the expression of the Hamiltonian you can see that $\hat{H}=\hat{H}_{1}+\hat{H}_{2}$. Also you have the following commutation relation $[\hat{H}_{1},\hat{H}_{2}]=0$. Thus, you can label the eigenstates as $|n_{1}n_{2},SM_{s}\rangle$. Where $|SM_{s}\rangle$ is the two particle total spin state.
Now, the ket corresponding to the ground state is $|n_{1}n_{2},SM_{s}\rangle=|00,00\rangle$. To see where the spin comes into play, we look at the first excited state. The first excited energy is $E_{1}=E(10)=E(01)$ with four possible kets:
For the singlet state $\left(\frac{1}{\sqrt{2}}|10\rangle+\frac{1}{\sqrt{2}}|01\rangle\right)|00\rangle$
For the triplet state $\left(\frac{1}{\sqrt{2}}|10\rangle+\frac{1}{\sqrt{2}}|01\rangle\right)|S=1,M_{s}=0,\pm1\rangle$
Having fermions, the antisymmetric wave function is
$$\psi=\frac{1}{\sqrt{2}}(\psi_{1}(x_{1})\psi_{2}(x_{2})-\psi_{1}(x_{2})\psi_{2}(x_{1}))$$
(there's a plus in your wave function and that is for integer spin particles). This wave function can be split into a spatial and spin part. Being an antisymmetric wave function, when the spatial part is symmetric the spin part is antisymmetric and vice versa.
$$\psi(x_{1},x_{2},M_{1},M_{2})=\left\{
\begin{array}{ll}
\psi^{S}(x_{1},x_{2})\chi^{A}(M_{1},M_{2})\\
\psi^{A}(x_{1},x_{2})\chi^{S}(M_{1},M_{2})
\end{array}\right.$$
where the spatial part is
$$\psi^{S}(x_{1},x_{2})=\frac{1}{\sqrt{2}}(\psi_{1}(x_{1})\psi_{2}(x_{2})+\psi_{1}(x_{2})\psi_{2}(x_{1}))$$
$$\psi^{A}(x_{1},x_{2})=\frac{1}{\sqrt{2}}(\psi_{1}(x_{1})\psi_{2}(x_{2})-\psi_{1}(x_{2})\psi_{2}(x_{1}))$$
and the spin part
$$\chi^{A}(M_{1},M_{2})=\frac{1}{\sqrt{2}}(\uparrow_{1}\downarrow_{2}-\uparrow_{2}\downarrow_{2}) \hspace{5mm} M_{s=}0,\hspace{5mm} S=0$$
$$\chi^{S}(M_{1},M_{2})=\left\{\begin{array}{ll}
\uparrow_{1}\uparrow_{2},\hspace{29mm} M_{s}=1\\
\frac{1}{\sqrt{2}}(\uparrow_{1}\downarrow_{2}+\uparrow_{2}\downarrow_{2}) \hspace{5mm} M_{s=0}\\
\downarrow_{1}\downarrow_{2} \hspace{29mm} M_{s}=-1
\end{array}\right \} , S=1$$
Best Answer
As mentioned in the comments, the wavefunction as posed is symmetric in the exchange between particles 1 and 2. This can be seen directly from the wavefunction, where to avoid confusion it is easier to take explicit spin components: \begin{align} \psi^{(0)}_{i_2,i_1,i_3}(x_2,x_1,x_3) &= \frac{1}{\sqrt{3!}} \begin{vmatrix} \psi_1(x_2) \langle i_2|{+}\rangle & \psi_1(x_1) \langle i_1|{-}\rangle & \psi_1(x_3) \langle i_3|{+}\rangle \\ \psi_1(x_2) \langle i_2|{-}\rangle & \psi_1(x_1) \langle i_1|{+}\rangle & \psi_1(x_3) \langle i_3|{+}\rangle \\ \psi_2(x_2) \langle i_2|{+}\rangle & \psi_2(x_1) \langle i_1|{+}\rangle & \psi_2(x_3) \langle i_3|{-}\rangle \\ \end{vmatrix} \\& \\ & = \frac{-1}{\sqrt{3!}} \begin{vmatrix} \psi_1(x_1) \langle i_1|{-}\rangle & \psi_1(x_2) \langle i_2|{+}\rangle & \psi_1(x_3) \langle i_3|{+}\rangle \\ \psi_1(x_1) \langle i_1|{+}\rangle & \psi_1(x_2) \langle i_2|{-}\rangle & \psi_1(x_3) \langle i_3|{+}\rangle \\ \psi_2(x_1) \langle i_1|{+}\rangle & \psi_2(x_2) \langle i_2|{+}\rangle & \psi_2(x_3) \langle i_3|{-}\rangle \\ \end{vmatrix} \\& \\ & = \frac{+1}{\sqrt{3!}} \begin{vmatrix} \psi_1(x_1) \langle i_1|{+}\rangle & \psi_1(x_2) \langle i_2|{-}\rangle & \psi_1(x_3) \langle i_3|{+}\rangle \\ \psi_1(x_1) \langle i_1|{-}\rangle & \psi_1(x_2) \langle i_2|{+}\rangle & \psi_1(x_3) \langle i_3|{+}\rangle \\ \psi_2(x_1) \langle i_1|{+}\rangle & \psi_2(x_2) \langle i_2|{+}\rangle & \psi_2(x_3) \langle i_3|{-}\rangle \\ \end{vmatrix} \\& \\ & = + \psi^{(0)}_{i_1,i_2,i_3}(x_1,x_2,x_3) \end{align}
Because of this symmetry under exchange, the wavefunction as written is not a suitable state for identical fermions.