Suppose $f:[0,1] \Rightarrow \mathbb{R}$ is continuous and $\int_0^x f(x)dx = \int_x^1 f(x)dx$. Prove that $f(x) = 0$ for all $x$

Here is another line of attack:

Since $\int_0^x f(t)dt = \int_x^1 f(t)dt$, we also have $\int_0^y f(t)dt = \int_y^1 f(t)dt$, so subtracting gives $\int_0^y f(t)dt-\int_0^x f(t)dt = \int_y^1 f(t)dt-\int_x^1 f(t)dt$, or $\int_x^y f(t)dt = \int_y^x f(t)dt = -\int_x^y f(t)dt$.

In other words, $\int_x^y f(t)dt = 0$ for all $x,y$.

Since $f$ is continuous, if $f(x_0) \neq 0 $ for some $x_0$, then there is some interval $[a,b]$ containing $x_0$ for which $|f(x)| \ge \frac{1}{2} |f(x_0)|$ for all $x \in [a,b]$. If follows that $\int_a^b f(t)dt \neq 0$, a contradiction.

The Fundamental Theorem of Calculus proof suggested in a comment by Peter Tamaroff is one short line, and one cannot do better.

Here is a more awkward proof that does not use the FTC. Suppose that $f(x)\ne 0$ for some $x$. Say for example that $f(a)=c\gt 0$ for some $a$. By continuity we can assume that $a$ is not $0$ or $1$. Then there is an interval $(a-\epsilon,a+\epsilon)$ contained in $(0,1)$ such that $f(x)\gt c/2$ in this interval.

Note that $\int_{a-\epsilon}^{a+\epsilon}\gt c\epsilon\gt 0$.

Let $x_1=a-\epsilon$, and $x_2=a+\epsilon$. Then if $\int_0^{x_1} f(t)\,dt=\int_{x_1}^1 f(t)\,dt$, we must have $\int_{0}^{x_2} f(t)\,dt \gt \int_{x_2}^1 f(t)\,dt$.

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}% \newcommand{\yy}{\Longleftrightarrow}$ $\ds{% \totald{}{x}\int_{0}^{x}{\rm f}\pars{\xi}\,\dd\xi = \totald{}{x}\int_{x}^{1}{\rm f}\pars{\xi}\,\dd\xi \quad\imp\quad {\rm f}\pars{x} = -{\rm f}\pars{x}\quad\imp\quad{\rm f}\pars{x} = 0}$.