Let $g(x) = f(x) - x^2$, then
$$\begin{align*} g(x+1) - g(x) &= f(x+1) - f(x) - (x+1)^2 + x^2 \\ &= (2x + 1) - (x^2 + 2x + 1 - x^2) \\ &= 2x + 1 - (2x + 1) \\ &= 0. \end{align*}$$
Thus, $g(x)$ is a periodic function with 1 as its period. On the other hand, as is given $|f(x)| \le 1$ when $x \in [0, 1]$, so
$$|g(x)| = |f(x) - x^2| \le |f(x)| + x^2 \le 1 + x^2 \le 2,$$
when $x \in [0, 1]$.
Therefore, the periodic function $g(x)$ satisfies $|g(x)| \le 2$ ($x \in \mathbb{R}$), thus arriving at
$$|f(x)| = |g(x) + x^2| \le |g(x)| + x^2 \le 2 + x^2 \quad (x \in \mathbb{R}),$$
as desired.