Goal. Show that in $L^2(\mathbb{R})$, $\{ \psi_i \}$ forms an orthonormal basis iff $\{ \widehat{\psi}_i \}$ forms an orthonormal basis.
Attempt.
In $L^2(\mathbb{R})$, we have the following polarization identity:
$$ \langle f, g \rangle = \langle \widehat{f}, \widehat{g} \rangle $$
which implies in particular that $\|f\| = \| \widehat{f} \|$ in $L^2(\mathbb{R})$.
This immediately implies that $\{ \psi_i \}$ forms an orthonormal set iff $\{ \widehat{\psi}_i \}$ forms an orthornomal set.
What remains to be shown is the basis portion of the theorem, that is that
$$ f(x) = \sum_{j,k} \langle f , \psi_{j,k} \rangle \psi_{j,k} \iff f(x) = \sum_{j,k} \langle f , \widehat{\psi}_{j,k} \rangle \widehat{\psi}_{j,k} $$
(Technically, I think we are taking "=" to mean equality in the $L^2$ sense.)
Question: How to show (3)?
An orthonormal set $\{ f_n \}$ is complete iff $$ \|f\|^2 = \sum_{n}|\langle f,f_n\rangle|^2, \;\;\; \forall f\in H. $$ This is one of the equivalents of completeness. Because $$ \|\hat{f}\|^2=\|f\|^2 = \sum_{n}|\langle f,f_n\rangle|^2 = \sum_n |\langle \hat{f},\hat{f}_n\rangle|^2,\;\;\; \forall \hat{f},\; f\in H, $$ and because $L^2 = \{ \hat{f} : f \in L^2 \}$, it follows that $$ \|f\|^2 = \sum_n |\langle f,\hat{f}_n\rangle|^2,\;\; \forall f\in L^2. $$ Therefore $\{ \hat{f}_n \}$ is a complete orthonormal basis of $L^2$ if $\{ f_n \}$ is a complete orthonormal basis. The same is true of the inverse transform as well, leading to the conclusion that $\{ \hat{f}_n \}$ is a complete orthonormal basis of $L^2$ iff $\{ f_n \}$ is a complete orthonormal basis of $L^2$.