What is the value of $f(2018)$ ?

Let $x,h \in \mathbb{R}$ . Then

$\begin{aligned}\lvert f(x+h)-f(x) \rvert &\leq \lvert x+h-x \rvert^2 = \lvert h \rvert^2\\\\ \implies 0 \leq \frac{\lvert f(x+h)-f(x) \rvert}{\lvert h \rvert} &\leq \lvert h \rvert\end{aligned}$ Letting $h \to 0$ we get $\lim_{h\to 0} \Bigg\lvert \frac{ f(x+h)-f(x)}{h} \Bigg\rvert = 0$ Therefore $f$ is differentiable with $f'(x)=0\ \ \text{on } \mathbb{R}$ , therefore $f$ is constant, so $f(2018)=f(0)=\boxed{1}$