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?
 A: Your proof looks  completely right to me. In fact, it's the one I was planning to write if the question hadn't included a proof. 
The only thing I'd change is to say that the integral, over the reduced interval, is at least $\frac{D}{2} 2\delta = \delta D > 0$. It's always nice to see a concrete estimate like that. 
I might also simplify a little and start out by observing that $\int_{[0, a]} f = 0$ for any $0 < a < 1$ because $f$ is nonnegative, so the integrals from $0$ to $a$ and $a$ to $1$ must be nonnegative, but they sum to zero, and hence are both zero. Further such reasoining shows that $\int_I f = 0$ for any subtinterval of $[0, 1]$. 
Once you've done that, you can stop your proof just before "This implies" by just saying "and now we've exhibited an interval over which the integral is positive, a contradiction."
A: Yes, I think there's no problem with your solution. We can say in short: since $f\ge 0$, its graph lies above $Ox$. If $f(x_0)>0$ for some $x_0$, then there is a neighborhood of $x_0$ such that for all $x_1<x_2$ in that neighborhood, we have $f(x_1)>0,f(x_2)>0$. Then the area under the curve on the segment $[x_1,x_2]$ is greater than $0$, which is impossible.
A: 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.
A: Let $F$ be a primitive of $f$ on $[a,b]$. ($F$ is differentiable and $F'=f$ ).
As $f$ is non-negative then $F$ is increasing.
As $\int_a^bf(t)dt=0$ then $F(b)-F(a)=\int_a^bf(t)dt=0$, that is $F(a)=F(b)$.
Conclusion : $F$ is increasing and $F(a)=F(b)$, hence $F$ is constant, so $f=F'=0$.
