$\int_{0}^{1}f(x)g(x)=0 \implies f(x)=0 \ \forall x \in [0,1]$ 
Let $f:[0,1]\to \mathbb{R}$ be a continuous function. If $\int_{0}^{1}f(x)g(x)=0$ for all continuous functions $g(x)$, then $f(x)=0 \ \forall x \in [0,1]$. I would like to know if my proof holds, please.

As the statement holds for every continuous function $g(x)$ on $[0,1]$, then it holds in particular for $g(x)=f(x)$. So, we have to show that $\int_{0}^{1}f^2(x)=0 \implies f(x)=0 \ \forall x \in [0,1]$.
First, as $f$ is continuous on $[0,1]$, $f^2(x)$ is continuous on $[0,1]$ as well. Therefore, $f^2(x)$ has a primitve $F(x)$ (which is differentiable on $[0,1]$). Thus,
$F(1)-F(0)=0\implies F(1)=F(0)$ and $F'(x)=f^2(x)\ge 0$.
As $F'(x)\ge 0$, then $F(x)$ is increasing. We will show now that $F(x)$ is constant.
Consider $0\le x\le 1$. As $F$ is increasing, then $F(0)\le F(x)\le F(1)=F(0)$. So, $F(x)=F(1)=F(0)=cste \ \forall x\in [0,1]$ .
We conclude that $F'(x)=cste'=0=f^2(x) \implies f(x)=0$ as wanted.

Just a suggestion : I guess you know that if $f\geq 0$ is continuous, then $\int_0^1 f=0\implies f=0$. Using that :
$$\forall g\in \mathcal C[0,1], \int_0^1 fg=0\implies \int_0^1 f^2=0\implies f=0.$$
