Demonstrating orthonormality for wave functions for ground state and excited state of He
First prove a well-known fact that any two spin orbitals $\psi_i$ and $\psi_j$ are orthogonal if the corresponding spatial orbitals from which they are constructed are themselves orthogonal. 


For two spin orbitals constructed from two different spatial orbitals orthogonality follows from orthogonality of spatial orbitals regardless of spin components.
For two spin orbitals constructed from the same spatial orbital orthogonality follows from orthogonality of spin functions.


Then you can work with spin orbitals to (re)establish a well-known result that for two Slater determinants differ in just one spin orbital, the overlap integral $\langle \Psi_0 \vert \Psi_1 \rangle$ is zero. 
$$
\Psi_0 = \frac{1}{\sqrt{2}}
\begin{vmatrix}
\psi_1(1) &amp; \psi_2(1) \\
\psi_1(2) &amp; \psi_2(2)
\end{vmatrix} \, ,
\quad
\Psi_1 = \frac{1}{\sqrt{2}}
\begin{vmatrix}
\psi_1(1) &amp; \psi_3(1) \\
\psi_1(2) &amp; \psi_3(2)
\end{vmatrix} \, ,
$$
where
$$
\psi_1(1) = \phi_1(1) \alpha(1) \, ,
\quad
\psi_2(1) = \phi_1(1) \beta(1) \, ,
\quad
\psi_3(1) = \phi_2(1) \beta(1) \, .
$$
Algebraically, it is very easy to show that,
$$
  \langle \Psi_0 \vert \Psi_1 \rangle =
  \frac{1}{\sqrt{2}} \frac{1}{\sqrt{2}} \langle \psi_1(1) \psi_2(2) - \psi_2(1) \psi_1(2) \vert \psi_1(1) \psi_3(2) - \psi_3(1) \psi_1(2) \rangle \\
  = \frac{1}{2} \Big(
    \langle \psi_1(1) \psi_2(2) \vert \psi_1(1) \psi_3(2) \rangle
  - \langle \psi_1(1) \psi_2(2) \vert \psi_3(1) \psi_1(2) \rangle \\
  - \langle \psi_2(1) \psi_1(2) \vert \psi_1(1) \psi_3(2) \rangle
  + \langle \psi_2(1) \psi_1(2) \vert \psi_3(1) \psi_1(2) \rangle \Big) \\
  = \frac{1}{2} \Big(
    \langle \psi_1(1) \vert \psi_1(1) \rangle
    \langle \psi_2(2) \vert \psi_3(2) \rangle 
  - \langle \psi_1(1) \vert \psi_3(1) \rangle
    \langle \psi_2(2) \vert \psi_1(2) \rangle \\
  - \langle \psi_2(1) \vert \psi_1(1) \rangle 
    \langle \psi_1(2) \vert \psi_3(2) \rangle
  + \langle \psi_2(1) \vert \psi_3(1) \rangle
    \langle \psi_1(2) \vert \psi_1(2) \rangle \Big) \\
  = \frac{1}{2} \Big(c_1 \cdot 0 - 0 \cdot 0 - 0 \cdot 0 + 0 \cdot c_1 \Big) = 0 \, .
$$
In the case when spin orbitals are normalized (to one) $c_1 = 1$, but it is not required here: the overlap integral will vanish for spin orbitals that are just orthogonal and not necessarily normalized (to one) as well.