Because if
$$\require{cancel}\hat{H}=\hat{H}_{1}+\hat{H}_{2}=\left[-\frac{\hbar^{2}}{2m}\nabla_{1}^{2}-\frac{e^{2}}{4\pi\varepsilon_{0}}\frac{1}{r_{1}}\right]+\left[-\frac{\hbar^{2}}{2m}\nabla_{2}^{2}-\frac{e^{2}}{4\pi\varepsilon_{0}}\frac{1}{r_{2}}\right]+\cancel{\color{red}{\frac{e^{2}}{4\pi\varepsilon_{0}}\frac{1}{r_{12}}}}$$
is your Hamiltonian and $\psi_{0}$ is the ground state of $\hat{H}_{1,2}$, i.e.
$$\hat{H}_{1}\psi_{0}\left(\vec{r}_{1}\right)=E_{0}\psi_{0}\left(\vec{r}_{1}\right)$$
$$\hat{H}_{2}\psi_{0}\left(\vec{r}_{2}\right)=E_{0}\psi_{0}\left(\vec{r}_{2}\right)$$
then you can argue that
$$\hat{H}\psi_{0}\left(\vec{r}_{1}\right)\psi_{0}\left(\vec{r}_{2}\right)=\left[E_{0}+E_{0}\right]\psi_{0}\left(\vec{r}_{1}\right)\psi_{0}\left(\vec{r}_{2}\right)$$
In other words, $\psi_{0}\left(\vec{r}_{1}\right)\psi_{0}\left(\vec{r}_{2}\right)$ is a solution with energy $2E_{0}$ which is of course the lowest possible.
A common hand-waving argument is that this choice satisfies the fact that
$$P\left(A\cap B\right)=P\left(A\right)P\left(B\right)$$
for independent events $A$ and $B$. Here $A=\color{blue}{{\rm electron\:1\:is\:at\:}r_{1}}$ and $B=\color{blue}{{\rm electron\:2\:is\:at\:}r_{2}}$ so
$$P\left(A\right)=\left|\psi_{0}\left(\vec{r}_{1}\right)\right|^{2}$$
$$P\left(B\right)=\left|\psi_{0}\left(\vec{r}_{2}\right)\right|^{2}$$
$$P\left(A\cap B\right)=\left|\psi_{0}\left(\vec{r}_{1}\right)\psi_{0}\left(\vec{r}_{2}\right)\right|^{2}=\left|\psi_{0}\left(\vec{r}_{1}\right)\right|^{2}\cdot\left|\psi_{0}\left(\vec{r}_{2}\right)\right|^{2}=P\left(A\right)P\left(B\right)$$
in agreement with the probabilistic interpretation of quantum mechanics.