The condition $|f(x) - f(y)| \le |x - y|$ is sufficient for the continuity of $f$. From $|x - y|^2 \le |f(x) - f(y)|$ we derive that $f$ is one-to-one, implying that $f$ is a strictly monotonic function. We may assume that $f$ is strictly increasing, because $f$ can be replaced by $-f$.
Set $x = 0$ and $y = 1$ to derive that $0 \le f(1) - f(0) \le 1$, hence $f(1) = f(0) + 1$.
For $x \ge y$ we get $f(x) - f(y) \le x - y$ or $y - f(y) \le x - f(x)$, implying that the function $g(x) = x - f(x) + f(0)$ is increasing.
Since $g(0) = g(1) = 0$, the function $g$ is the identically null. Consequently, the functions are $f_a^{\pm}$ given by $f_a^{\pm}(x) = \pm x + a$, with $a \in \mathbb{R}$.

Alternative solution:
For $x = 0$ and $y = 1$ we get $|f(1) - f(0)| = 1$. For $x \in [0, 1]$ we obtain $1 = |f(1) - f(0)| \le |f(1) - f(x)| + |f(x) - f(0)| \le |1 - x| + |1 - x| = 1 - x + x - 0 = 1$.
Thus $f(x)$ lies between $f(0)$ and $f(1)$; moreover, we have $|f(1) - f(x)| = 1 - x$ and $|f(x) - f(0)| = x$. It follows that points $(0, f(0))$, $(x, f(x))$ and $(1, f(1))$ are collinear, hence $f(x) = f(0) \pm x$.