$\displaystyle \int_{0}^{1}{|f|}=0\Rightarrow f=0 $
given f is continuous on [0,1]
you said continuous on [0,1], that is not enough, as the function $\displaystyle f: \mathbb{R} \mapsto \mathbb{R}$ defined by
$\displaystyle f(x) = \begin{cases} 0 &\mathrm{if}\,\,0\leq x \le1 \\ 1& \mathrm{otherwise}\end{cases}$
is continuous on [0,1] but is not identically zero, yet $\displaystyle \int_0^1 |f(x)|~dx = 0$
If $\displaystyle f$ is continuous on $\displaystyle \left[ {0,1} \right]$ so is $\displaystyle \left| f \right|$.
Suppose that $\displaystyle \left( {\exists c \in \left[ {0,1} \right]} \right)\left[ {\left| {f(c)} \right| > 0} \right]$ then $\displaystyle \left( {\exists \left[ {u,v} \right] \subseteq \left[ {0,1} \right]} \right)\left[ {\left( {\forall z \in \left[ {u,v} \right]} \right)\left[ {\left| {f(z)} \right| > 0} \right]}\right]$.
What can we say about $\displaystyle \int\limits_u^v {\left| {f(z)} \right|dz} $.
Do you see the contradiction in that?