= Solution
For $\varepsilon>0$, insert a <Gaussian function> into the proposed <inverse Fourier transform>:
$$
I_\varepsilon(x)=\frac1{2\pi}\int_{\mathbb R}\widehat\phi(\xi)e^{ix\xi}e^{-\varepsilon\xi^2/2}\,d\xi.
$$
The <Fourier bounds for Schwartz functions> show that $\widehat\phi$ is an <integrable function>, so <dominated convergence> gives
$$
I_\varepsilon(x)\longrightarrow\frac1{2\pi}\int_{\mathbb R}\widehat\phi(\xi)e^{ix\xi}\,d\xi.
$$
On the other hand, <Fubini's theorem> applies because $\|\phi\|_1\int e^{-\varepsilon\xi^2/2}\,d\xi<\infty$. The given <Fourier transform of a Gaussian>, rescaled by $\sqrt\varepsilon$, yields
$$
\begin{aligned}
I_\varepsilon(x)
&=\int_{\mathbb R}\phi(y)\left(\frac1{2\pi}\int_{\mathbb R}e^{-\varepsilon\xi^2/2}e^{i(x-y)\xi}\,d\xi\right)dy\\
&=\int_{\mathbb R}\phi(y)\frac{e^{-(x-y)^2/(2\varepsilon)}}{\sqrt{2\pi\varepsilon}}\,dy
=\frac1{\sqrt{2\pi}}\int_{\mathbb R}\phi(x-\sqrt\varepsilon z)e^{-z^2/2}\,dz.
\end{aligned}
$$
The last <integrand> is bounded by $\|\phi\|_\infty e^{-z^2/2}/\sqrt{2\pi}$. Since $\phi$ is continuous, another application of <dominated convergence> makes this tend to $\phi(x)$. Equating the limits proves the <Fourier inversion theorem>:
$$
\boxed{\phi(x)=\frac1{2\pi}\int_{\mathbb R}\widehat\phi(\xi)e^{ix\xi}\,d\xi\qquad(x\in\mathbb R).}
$$
The <Gaussian functions> above form an <approximate identity>; no <Fourier inversion> was used in evaluating their <integrals>.
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