Prove or disprove that if $f:\Bbb R\to\Bbb R$ is such that $|f(x)-f(y)|<|x-y|$ for all different real numbers $x$, $y$, then $f(x)=x$ has a unique solution.
Firstly, there can't be $f(a)=a$ and $f(b)=b$ for different real numbers $a$, $b$, since $$|a-b|>|f(a)-f(b)|=|a-b|$$ is absurd.
So we assume that $f(x)\ne x$ for all real numbers $x$ and see whether we arrive at another contradiction (proving the statement) or manage to find a counterexample (disproving the statement).
Let $A=\{x\mid f(x)>x\}$, $B=\{x\mid f(x)<x\}$, then $A$, $B$ form a partition of $\Bbb R$. Assuming there exists $a\in A$ and $b\in B$ such that $a>b$, then $f(a)>a>b>f(b)$, and so $$|f(a)-f(b)|=f(a)-f(b)>a-b=|a-b|,$$ contradiction! So for all $a\in A$ and $b\in B$, we have $a<b$.
I am stuck here. After some thinking, perhaps whether $f(x)$ is greater or less than $x$ is not so important...?
Counter-example: The function $f(x) = \frac{\vert x\vert}{1 + \vert x\vert}x + 1$ has no fixed point, however satisfies $\vert f(x) - f(y)\vert < \vert x - y\vert$.