I should prove that given a function $f \in C_0$, then if $\int_{\mathbb{R}} \int_{\mathbb{R}} f(x-t)(\phi_{tt} - \phi_{xx})dxdt =0$ for every $\phi \in C_{0}^{\infty}$, then $f=0$.
I had thought that integrating by parts (with respect to $t$) would give us something like $\int_{\mathbb{R}} \int_{\mathbb{R}} f(x-t)(\phi_{tt} - \phi_{xx})dxdt = \int_{\mathbb{R}} \int_{\mathbb{R}} F(x-t)(\phi_{ttt} - \phi_{xxt})dxdt$, with $F$ a primitive of $f$. Now if I take $\phi= \psi(2x +2t)+ \psi(x-2t)$ with $\psi \in C_{0}^{\infty}$ I should get $\phi_{ttt} - \phi_{xxt}= \psi(x-2t)$, ignoring costants multiplicating and derivatives, since I am working with arbitrary $C^{\infty}$ functions. So I can now study the problem $\int_{\mathbb{R}} \int_{\mathbb{R}} F(x-t) \psi(x-2t)dxdt =0$. Integrating by parts again should give something like (again ignoring integrals and derivatives for $\psi$) $\int_{\mathbb{R}} \int_{\mathbb{R}} f(x-t)\psi(x-2t)dxdt =0$ and now I should be able to use the fundamental theorem of calculus of variations to conclude.
Is this proof right? Is there a nicer way of proving the statement? Thanks for the help!
There are multiple issues with your approach, for example a smooth compactly supported function $\phi$ cannot be written as $\phi= \psi(2x +2t)+ \psi(x-2t)$, as the sum on the right is not compactly supported (consider the line $x=2t$, for example).
Also, the claim is false. Change the variables $u=x-y$, $v=x+t$ and note that $\phi_{tt}-\phi_{xx} = \phi_{uv}$. The assumption now becomes $$ \iint_{\mathbb{R}^2} f(u)\phi_{uv}(u,v)\,du\,dv = 0 \tag1 $$ But (1) holds for every $f$, because the integral over $v$ is $$ \int_{-\infty}^\infty f(u)\phi_{uv}(u,v) \,dv = f(u) \phi_u(u, \cdot)\bigg|_{-\infty}^\infty = 0 $$ So, any continuous compactly supported function works.
Since the wave operator is self-adjoint (one can throw it to the other function when both are smooth enough), the problem asserts that a function of the form $f(x-t)$ with $f\in C_0$ cannot be a solution of the wave equation in the distributional sense. But it can; indeed any such function is a solution, the general solution being $f(x-t)+g(x+t)$.