We claim $f(x)=0$ for all $x$ is the only such function. Suppose there is an $f$ not identically zero, and let $f(x_0):=c\ne 0$. Define the function $g(x):=f(x)+x$. Then, $f(g(x)) = 2f(x)$. Define next a sequence $x_k$ such that $x_k=g(x_{k-1})$ for $k\ge 1$. Clearly, $f(x_1)=f(g(x_0)) = f(f(x_0)+x_0)=2f(x_0)=2c$. Iterating this, $f(x_2) = f(g(x_1))=f(f(x_1)+x_1)=2f(x_1)=4c$ and by induction, $f(x_k) =2^k c$. As $c\ne 0$, this contradicts with the boundedness of $f$ for sufficiently large $k$.

f(f(x)+y)=f(x)+f(y) let y = 0 f(f(x)) = f(x)+f(0) f(a)=a+f(0) where f(0) is a constant f(x)=x+c

StarLex1 wrote: f(f(x)+y)=f(x)+f(y) let y = 0 f(f(x)) = f(x)+f(0) f(a)=a+f(0) where f(0) is a constant f(x)=x+c You can't conclude $f(x)=a$ without proving surjectivity