Let $I_k=\left [ k,k+1 \right ]$ where $k$ is a postive integer. Fist of all take a $x_0 \in I_0$ we find easly using the condition that $\left |f(x_0) \right |\leq1\leq x_0^2+2$ now we have $\left |f(x_0+1) \right |=|f(x_0)+2x_0+1|\leq |f(x_0)|+2x_0+1\leq (x_0+1)^2+2$ And so mapping $I_0$ yields that the relation is satisfied for $I_1$ Doing the same we get that teh relation is satisfied for $I_2,I_3,.......$ and so for all $x\geq 0.$ $|f(-x_0)|=|f(-x_0+1)+2x_0-1|$ If $x_0\geq \frac{1}{2}$ $|f(-x_0)|\leq |f(-x_0+1)|+2x_0-1\leq (-x_0+1)^2+2 x_0 -1+2 \leq x_0^2+2$ (because $-x_0+1\in I_0$ ) If $x_0 \leq \frac{1}{2}$ $|f(-x_0)|\leq |f(-x_0+1)|-2x_0+1\leq 2-2x_0\leq x_0^2+2$ (this last inequality is true because $x_0\geq 0.$ ) And by the same argument we reach $I_{-1} ,I_{-2} ....$ finally we get that the relation is satisfied for all reals .

We call $g(x)=f(x)-x^2$, so $g(x+1)=f(x+1)-(x+1)^2, g(x+1)-g(x)=f(x+1)-f(x)-(2x+1)=0, g(x+1)=g(x)$ So we can get that $g(x)$ is a periodic function with period $1$ over reals. As $x\in [0,1], |f(x)|\leq 1$, so $|g(x)|=|f(x)-x^2|\leq |f(x)|+|x^2|\leq 2, |f(x)|=|g(x)+x^2|\leq |g(x)|+|x^2|\leq 2+x^2$ as desired