Let $f,g\in L^2(I)$. Prove that if $$\int_If\phi dm=\int_Ig\phi dm$$ for all $\phi\in C(I)$, then $f=g$ a.e over $I$, which is a finite interval.
Attempt
By hypothesis $\int_If\phi dm=\int_Ig\phi dm$.
Then considering this result $\int_Efdm=\int_Egdm\implies f=g, a.e$
implies $f\phi=g\phi$ a.e
From here, how could I 'remove' $\phi$ to have only $f=g$?
Can someone please help me?
Your attempt has errors. You cannot conclude that $f \phi =g\phi$ a.e. because the given equation holds only for the given interval. It is not given that the integrals over any measurable set $E$ are equal. This makes the result more involved.
As commented above, you have to assume that $I$ is a finite interval.
Here are some hints: given any closed interval $[a,b]$ contained in $I$ we can find continuous functions $\phi_n$ such that $ 0 \leq \phi_n \leq 1$ and $\phi_n(x) \to 1$ if $ x\in [a,b]$ and $\phi_n(x) \to 0$ if $ x\notin [a,b]$. Applying the hypothesis and letting $n \to \infty$ we get $\int_a^{b} f(x)dx=\int_a^{b} g(x)dx$ (1) for all such intervals $[a,b]$. We are still far away from the conclusion because we cannot immediately replace the interval $[a,b]$ by an arbitrary Borel set. One result that gives $f=g$ a.e. immediately is Lebesgue's Theorem. If you divide both sides of (1) by $b-a$ and let $b \to a$ you get $f=g$ almost everywhere from this theorem.
Alternative proof: the proof is much simpler if you know that $L^{2}$ can approximated by continuous functions. Let $\phi_n$ be continuous functions converging in $L^{2}$ to $f-g$. Then you can pass to the limit in $\int_I(f-g)\phi_n=0$ to get $\int (f-g)^{2}=0$ which implies $f=g$ a.e..