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
$$\chi^{A}(M_{1},M_{2})=\frac{1}{\sqrt{2}}(\uparrow_{1}\downarrow_{2}-\uparrow_{2}\downarrow_{2}) \hspace{5mm} M_{s=}0,\hspace{5mm} S=0$$
$$\chi^{S}(M_{1},M_{2})=\left\{\begin{array}{ll}
\uparrow_{1}\uparrow_{2},\hspace{29mm} M_{s}=1\\
\frac{1}{\sqrt{2}}(\uparrow_{1}\downarrow_{2}+\uparrow_{2}\downarrow_{2}) \hspace{5mm} M_{s=0}\\
\downarrow_{1}\downarrow_{2} \hspace{29mm} M_{s}=-1
\end{array}\right \} , S=1$$
People often say "quantum indistinguishable particles that are very far-separated behave like distinguishable particles," but this is a bit misleading. It would be more accurate to say "quantum indistinguishable particles that are very far-separated behave like classical indistinguishable particles." The difference is subtle but important.
The problem is in your sentence "if I look at $x_1$, $x_2$, with $x_1 \in \text{supp}(\psi_1(\cdot))$ and $x_2 \in \text{supp}(\psi_2(\cdot))$ ...". There's no possible experiment that says "I've detected particle $x_1$ at location $x$" - only experiments that say "I've detected a particle at location $x$." So the probability amplitude that you measure a particle at point $x$ and another particle at point $y$ is
$$\langle x, y | \frac{1}{\sqrt{2}}(|\psi_1 \psi_2 \rangle + | \psi_2 \psi_1 \rangle) = \frac{1}{\sqrt{2}}(\psi_1(x) \psi_2(y) + \psi_2(x) \psi_1(y))$$
and the actual probability is the norm-squared
$$P(x, y) = \frac{1}{2} \left( |\psi_1(x)|^2 |\psi_2(y)|^2 + |\psi_2(x)|^2 |\psi_1(y)|^2 \right),$$
where the cross-terms are all zero because the states $|\psi_1\rangle$ and $|\psi_2\rangle$ are orthogonal. This is indeed the result for classical indistinguishable particles, and $\int P(x, y)\ dx\, dy = 1$ as it should (unlike your expression).
The only things that differ from the generic quantum case are (a) you can just use the normalization factor of $1/\sqrt{2}$ and (b) when you norm-square the amplitude out to an actual probability density, you can ignore the cross-terms.
If you want to work classically right from the beginning, then you don't symmetrize the ket at all, but work with the ket $|\psi_1 \psi_2 \rangle$ instead. Then your normalization issue doesn't come up.
Best Answer
The simple product $\psi_1(x_1)\psi_1(x_2)$ is already symmetric under $x_1 \leftrightarrow x_2$. So it is not necessary to symmetrize the wave function by adding in the term you get when you switch $x_1$ and $x_2$.