Prove that $\,f=0$ almost everywhere.

Define $g(x)=\int_0^x f(t)dt$. Then $g$ satisfies $|g(x)-g(y)|\leq (x-y)^2$. This implies at once that $g$ is continuous, differentiable with $g'(x)=0$. Therefore $g$ is constant and $g(0)=0$ implies that $g(x)=0$.

This means that $f$ is integrable with $\int_I f=0$ for every interval $I \subset [0,1]$ and this extends to $\int_B f=0$ for every Borel measurable set $B \subset [0,1]$. Consider now the sets $A_n = \{ |f| \geq \frac{1}{n}\}$. Then the sets $A_n$ are Lebesgue measurable and we have $\mu(A_n)=\sup\limits_{K \subset A_n, \text{ compact} }\mu(K)=0$ (since compact sets are Borel measurable; $\mu$ is the Lebesgue measure).

Therefore $$\mu(\{f\neq 0\})=\mu(\bigcup_n A_n) =\lim_{n\to \infty} \mu(A_n)=0 $$ so $f=0$ almost everywhere.

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Fact. For every $\varepsilon>0$, there exists a $\delta>0$, such that $m(E)<\delta$ implies that $\int_E|f|dx<\varepsilon$.

This is due to the fact that $f$ is integrable.

We have that $$ \{x: f(x)\ne 0\}=\bigcup_{k\in Z} \{x: f(x)\in (2^k,2^{k+1})\}\cup \bigcup_{k\in Z} \{x: f(x)\in (-2^{k+1},-2^{k})\}=\bigcup_{k\in Z}A^+_k\cup\bigcup_{k\in Z}A^-_k. $$ It suffices to show that $$ m(A_k^+)=m(A_k^-)=0, \quad\text{for all}\,\,\, k\in \mathbb Z. $$ Assume that $m(A_k^+)=a>0$. For every $\varepsilon>0$, there exists an open set $U$, such that $A_k^+\subset U$ and $m(U\smallsetminus A_k^+)<\varepsilon$. Then $$ \int_U f\,dx=\int_{A_k^+}f\,dx+\int_{U\smallsetminus A_k^+}f\,dx. $$ Clearly, $\int_{A_k^+}f\,dx\ge 2^ka$. Using the Fact above, we can choose $\varepsilon$ small enough, so that $\int_{U\smallsetminus A_k^+}|f|<2^{k-1}a$, and therefore $$ \int_{U\smallsetminus A_k^+}f\,dx\ge-2^{k-1}a, $$ and hence $\int_U f\,dx\ge2^{k-1}a.$ But, as $U$ is open it can be written as a union of disjoint open intervals: $U=\cup_{n\in\mathbb N}I_n$, and $$ 0<\int_U f\,dx=\sum_{n\in\mathbb N}\int_{I_n}f\,dx, $$ which means that for some interval $I_n=(c,d)$ we should have $$ \int_c^d f\,dx>0. $$ Using now the assumption, for every $n\in\mathbb N$, we have $$ \int_c^d f\,dx=\sum_{k=1}^n\int_{c+\frac{(k-1)(d-c)}{n}}^{c+\frac{k(d-c)}{n}} f\,dx\le n\cdot \left(\frac{d-c}{n}\right)^2=\frac{(d-c)^2}{n}, $$ which of course implies that $\int_c^d f\,dx=0$, which is a contradiction. Thus $f=0$ a.e.

The Lebesgue differentiation theorem states that if $f$ is integrable, then $f(x) = \lim_{\epsilon \downarrow 0} {1 \over 2\epsilon} \int_{x-\epsilon}^{x+\epsilon} f(t) dt$ for ae. [$m$] $x \in [0,1]$. Since $| {1 \over 2\epsilon} \int_{x-. \epsilon}^{x+\epsilon} f(t) dt | \le { (2 \epsilon)^2\over 2 \epsilon} $, we see that $f(x) = 0$ ae. [$m$] $x \in [0,1]$.