Please note that 1) My analysis background is weak, so things that may be "obvious" are likely not obvious to me. 2) My question is not "how to prove Plancharel's". I know there is much information on this available online.
In my course we have proven that Plancherel's holds for Schwartz functions (rapidly decreasing functions), that is $f \in \mathcal{S}(\mathbb{R})$ then $\lVert \hat{f} \rVert_{2} = \sqrt{2\pi} \, \lVert{f}\rVert_{2}$, where the constant is just because of our normalization of the Fourier Transform.
I would like to prove that more generally this result holds for functions in $L^{1} \cap L^{2}$. We have proven in class that $\mathcal{S}(R)$ is dense in $L^{1} \cap L^{2}$ so for some $f \in L^1 \cap L^2$ I know I could choose some sequence of Schwartz functions, $(\phi_j)_{j \in \mathbb{N}}$ such that $$\phi_j \to f \tag{in $L^1$ and $L^2$ norm?},$$ then I also know by Riemann-Lesbegue that this implies $$\hat{\phi_j} \to \hat{f} \tag{uniformly, in $\infty$ norm?}$$
But then how can I use this information (and possibly other facts) to show that Plancherel's holds for $f$ as well?
Use Cauchy sequences ; $||\phi_j -\phi_k||_2 \leq ||\phi_j -f||_2 +||f -\phi_k||_2 \to 0$. Hence $||\hat \phi_j - \hat \phi_k ||_2 \to 0$. Now $||\phi_j -f||_1 \to 0$ and this implies $\hat \phi_j - \hat f \to 0$ pointwise because $|\hat {h}| \leq ||h||_1$ (upto a factor of $\sqrt {2\pi}$) for any $L^{1}$ function h. The fact that $\{ \hat {\phi_j}\}$ is Cauchy in $L^{2}$ implies it converges in $L^{2}$ and since this sequence converges pointwise to $\hat f$ it follows that $\hat \phi_j \to \hat f$ in $L^{2}$. ( $L^{2}$ convergence implies pointwise convergence almost everywhere for a subsequence so the pointwise and $L^{2}$ limits coincide). Now just take limits in $||\hat \phi_j||_2 =\sqrt {2\pi} ||\phi_j||_2$.