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Define $F(x):= \int_0^xf(t)\,dt$ for $x \in [0,1]$. Then the second integral tells us $$ \begin{aligned} 0 = \int_0^1xf(x)\,dx \, & = \, xF(x)\Bigr|_0^1-\int_0^1F(x)\,dx \\ & = 1\,F(1) - 0\,F(0) - \int_0^1F(x)\,dx \\ & = \int_0^1f(x)dx - \int_0^1F(x)\,dx \\ & = -\int_0^1F(x)\,dx. \end{aligned} $$

By the mean value theorem and continuity of $F$, which follows from continuity of $f$, this tells us that there is $c \in (0,1)$ such that $F(c)=0$. That is, $$ \int_0^cf(x)\,dx = 0. $$ It follows that $$ \int_c^1f(x)\,dx = \int_0^1f(x)\,dx-\int_0^cf(x)\,dx = 0 - 0 = 0 $$ as well.

We then apply the mean value theorem two more times, using continuity of $f$, for the intervals $[0,c]$ and $[c,1]$ to find $a \in (0,c)$ and $b \in (c,1)$ respectively such that $f(a)=f(b)=0$.