If $f$ has a double zero at a point $z$ then not only $z=\sin z$ there
but also the derivatives must be the same, so that $1=\cos z$ there.
Letting $w=e^{iz}$ we obtain the equation $w+\frac1w=2i$ from which it
follows $w=i(1\pm\sqrt2)$ is pure imaginary.  But then
$z=\frac{w-\frac1w}{2i}$ is real, a contradiction. [The computational error is corrected in the *comments* and the other *answer*.]  Since $f$ is a transcendental function, it must hit every value infinitely many times, including the value $0$.  Since the zeros other than $z=0$ are simple, the square root cannot be an entire function.  The function $g$ does exist as a real analytic function; its radius of convergence at $0$ is probably the distance to the nearest zero of $f$, but I haven't checked this.