I have found a proof to the following theorem which is a fair bit shorter than the proof in my notes. I would be very grateful if someone could tell me whether this way works or whether I've made an error somewhere. Many thanks.
NOTE: If $f \in L^2(\mathbb{R})$, define its Fourier transform as follows:
There exists a sequence of Schwartz functions $(f_k)$ such that $\|f-f_k\|_2\to 0$ as $k\to\infty$. In particular, the sequence $(f_k)$ is Cauchy and so, by Parseval's identity for Schwartz functions, the sequence $(\,\widehat{f_k}\,)$ of Schwartz functions is also Cauchy and, since $L^2(\mathbb{R})$ is complete, it has a limit, say $g \in L^2(\mathbb{R})$. We define this $g$ to be the $L^2$ Fourier transform of $f$ and denote it by $\widetilde{f\,}$.
Theorem. Let $f \in L^2(\mathbb{R})$. For each $R>0$ define $f_R \in L^2(\mathbb{R})$ by $$f_R(x) = f(x)\chi_{[-R,R]}(x)$$ where $\chi_{[-R,R]}$ denotes the indicator function of $[-R,R]$. Then $$\|\widetilde{f\,}-\widehat{f_R}\|_2\to 0 \text{ as } R\to \infty$$ where $\widetilde{f\,}$ denotes the $L^2$-Fourier transform of $f$.
Proof. Let $\varepsilon>0$ be given and let $(f_k)$ be a sequence of Schwartz functions such that $$\|f-f_k\|_2\to 0 \text{ as } k\to\infty$$ so that, by definition, we have $$\|\widetilde{f\,}-\widehat{f_k}\|_2\to 0 \text{ as } k\to\infty$$ For any $k \in \mathbb{N}$ we have $$\|\widetilde{f\,}-\widehat{f_R}\|_2\leq\|\widetilde{f\,}-\widehat{f_k}\|_2+\|\widehat{f_R}-\widehat{f_k}\|_2$$ But, by Parseval's identity, we have that $$\|\widehat{f_R}-\widehat{f_k}\|_2=\|f_R-f_k\|_2\leq\|f_R-f\|_2+\|f-f_k\|_2$$ Combining the two inequalities yields $$\|\widetilde{f\,}-\widehat{f_R}\|_2\leq\|\widetilde{f\,}-\widehat{f_k}\|_2+\|f_R-f\|_2+\|f-f_k\|_2$$ Since this is true for all $k \in \mathbb{N}$ we choose $k$ so large that $$\|\widetilde{f\,}-\widehat{f_k}\|_2+\|f-f_k\|_2\leq \varepsilon/2.$$ Clearly, for large enough $R$, we have $$\|f_R-f\|_2<\varepsilon/2$$ and so, the result follows. Q.E.D.