I want to prove it through the hint given in the notes available online(link provided below). It says first prove that if $f\in C_c^2(\mathbb{R})$, then $\hat{f}\in L_1(\mathbb{R})$; and hence conclude range of fourier transform on $L_1(\mathbb{R})$ is dense in $C_0(\mathbb{R})$. Conclusion part is okay, but I am unable to prove that if $f\in C_c^2(\mathbb{R})$, then $\hat{f}\in L_1(\mathbb{R})$. For those who are not familiar with the terminology, let me recap: $C_c^2(\mathbb{R})$ denotes the collection of continuous functions with compact support which are square integrable.
Link of the notes: http://people.math.gatech.edu/~heil/handouts/chap1.pdf
see page $\textbf{72}$, Exercise $\textbf{1.100}$
For $x\ne 0,$
$$\hat f (x) =\int_\infty^\infty f(t)e^{-ixt}\, dt =1/(ix)\int_\infty^\infty f'(t)e^{-ixt}\, dt = 1/(ix)^2\int_\infty^\infty f''(t)e^{-ixt}\, dt.$$
I've integrated by parts twice; each time the boundary terms disappear because of compact support. The last expression is $1/x^2$ times a function in $C_0.$ We already know $\hat f$ is bounded. The above shows it is bounded by a constant times $1/x^2$ for $|x|$ large. Hence $\hat f \in L^1.$