Suppose $\int_{[a,b]}f=0,\text{ then }f(x)=0 \forall x\in[a,b]$ 
Let $a<b$ be real numbers. Let $f:[a,b]\to\mathbb{R}$ be a continuous non-negative function. Suppose $\int_{[a,b]}f=0,\text{ then }f(x)=0 \forall x\in[a,b]$ 

Proof:
Suppose for the sake of contradiction $\exists x_0\in[a,b] \text{ such that } f(x_0)= D> 0$
Since the function is continuous at $x_0$, we have
$$\forall \epsilon>0, \exists \delta>0 \text{ such that } |f(x)-f(x_0)|\leq\epsilon\text{ whenever } x\in[a,b]\cap(x_0-\delta,x_0+\delta)$$
choose $\delta=\delta_1$ such that $\epsilon=D/2$, then we can say:
$$-D/2 \leq f(x)-f(x_0)\leq D/2$$
$$-D/2 \leq f(x) - D$$
$$f(x)\geq D/2>0$$
This implies that $\int_{[a,b]\cap(x_0-\delta_1,x_0+\delta_1)}f >0$
Now since $\int_{[a,b]} f = 0$. we have
$$\int_{[a,b]}f=\int_{[a,x_0-\delta_1]}f+ \int_{(x_0-\delta_1, x_0+\delta_1)} f + \int_{[x_0+\delta_1,b]}f=0$$
$$\int_{[a,x_0-\delta_1]}f+\int_{[x_0+\delta_1,b]}f<-D/2,$$ but this is a contradiction since $f(x)\geq 0$ and hence $\int_{[a,x_0-\delta_1]}f+\int_{[x_0+\delta_1,b]}f\geq 0$
Is my proof correct?

This isn't exactly what you're asking for, but this is how I would approach this problem:
If $f$ is Riemann integrable, then it is Lebesgue integrable and the two integrals are equal. So if $\int_{[a,b]}f\mathrm dm=0$ where $m$ is Lebesgue measure, then $f=0$ almost everywhere (since $f$ is nonnegative). Since $f$ is continuous, it follows that $f$ is identically zero.
