A functional equation with f(2x)

1. From $f(2y)=f(y)$ with $y=\frac x2$ we get $f(x)=f\!\left(\frac x2\right)$, and by induction $f(x)=f\!\left(\frac x{2^n}\right)$ for every $n$. As $n\to\infty$, $\frac x{2^n}\to0$ and by continuity at $0$, $f\!\left(\frac x{2^n}\right)\to f(0)$. Since the left side does not depend on $n$, $f(x)=f(0)$ for all $x$.

2. Put $g(x)=f(x)-x$. Then
$$g(2x)-g(x)=\bigl(f(2x)-2x\bigr)-\bigl(f(x)-x\bigr)=\bigl(f(2x)-f(x)\bigr)-x=0.$$
So $g$ satisfies $g(2x)=g(x)$ and is continuous at $0$. By part 1, $g$ is a constant $c$. Hence $f(x)=x+c$.