Why does this integral inequality hold? The original question was: 

Let $f$ be a continuous function on $[a,b]$ and suppose that for every $a_1$ and $b_1$ in this interval:
  $$\int_{a_1}^{b_1}f(x)\text{d}x=0$$
Show that $f(x)=0$. 

So the answer given proves that there exists a $\delta>0$ such that $\displaystyle f(x)>\frac{f(c)}{2}$, with $c\in (a,b)$. 
Then the proof says that by picking $\delta$ small enough we can have $(c,c+\delta)$ or $(c-\delta,c)$ lies within the interval, which I understand.
Thus we can have: $$\int_c^{c+\delta}f(x)dx>\frac{f(c)}{2}\delta>0$$
Can someone please explain the last inequality for me please? I know from MVT that there exists a $c\in (a,b)$ such that $\displaystyle f(c)(b-a)=\int_a^b f(x)dx$, but not sure how I can go about using this to prove the above inequality.
 A: Let $\varepsilon>0$. Choose $\delta>0$ such that $|x-y|\leqslant\delta\implies |f(x)-f(y)|<\varepsilon$. Since $[a,b]$ is compact, there exist $a=x_1<x_2<\cdots<x_n=b\in [a,b]$ such that $$[a,b]\subset\bigcup_{j=1}^n \left(x_j-\frac\delta2,x_j+\frac\delta2\right).$$
Suppose there exists $x^\star\in[a,b]$ such that $|f(x^\star)|>0$. Then $$x^\star\in\left[x_{j^\star}-\frac\delta2,x_{j^\star}+\frac\delta2\right]$$ for some $1\leqslant j^\star\leqslant n$. If $y\in\left[x_{j^\star}-\frac\delta2,x_{j^\star}+\frac\delta2\right]$ then $$0<|f(x^\star)|\leqslant |f(x^\star)-f(y)|+|f(y)|<\varepsilon+|f(y)|, $$ so $0<|f(x^\star)|\leqslant |f(y)|$. Assume $WLOG$ that $f(x^\star)>0$, then $0<f(x^\star)\leqslant |f(y)|$ and continuity of $f$ implies that $f(y)>0$. But this means that $f$ is strictly positive on $\left[x_{j^\star}-\frac\delta2,x_{j^\star}+\frac\delta2\right]$, and so 
$$\int_{x_{j^\star}-\frac\delta2}^{x_{j^\star}+\frac\delta2} f(x)\ \mathsf dx \geqslant \delta f(x^\star)>0, $$ a contradiction.
