We know period of $\sin x$ is $2π.$ So period of $\sin cx$ will be $\frac{2π}{|c|}.$ Therefore period of $(\sin x+\sin cx)$ is:
$\text{LCM of }\left(2π, \frac{2π}{|c|}\right)=\frac{\text{LCM of }(2π, 2π)}{\text{HCF of }(1,|c|)}.$
Now if $c\in\mathbb R\smallsetminus\mathbb Q,$ then HCF of $1$(rational) and $|c|$(irrational) is not possible. But since $(\sin x+\sin cx)$ is periodic, so $c$ must be rational.
Conversely if $c\in\mathbb Q,$ then period of $(\sin x+\sin cx)$ is:
$\text{LCM of }\left(2π, \frac{2π}{|c|}\right)=\frac{\text{LCM of }(2π,2π)}{\text{HCF of }(1, |c|)}, \text{ which is possible as }|c|\in\mathbb Q.$
So the period of $f(x)=\sin x+\sin cx$ is $2π.$ Which is correct since $f(x+2π)=f(x).$
Hence the statement follows.
This was my approach. But I, personally, don't like this approach that much. So is there any other direct or obvious prove than this? Please suggest..
Suppose $f(x)$ is periodic with period $L$. This means $f(x+L)=f(x)$ for all real $x$. Then, $g_a(x)=f(x)+f(x+a)$ (with some constant $a$) is periodic with period $L$, too: $g_a(x+L)=f(x+L)+f(x+a+L)=f(x)+f(x+a)=g_a(x).$ Now, consider $f(x)=\sin x+\sin cx.$ If it is periodic with period $L$, so is $$g_\pi(x)=\sin cx + \sin(cx+c\pi)=2\sin(cx+c\pi/2)\cos(c\pi/2)$$ (since $\sin x + \sin(x+\pi)=0. $) This can be $0$, if $\cos c\pi/2=0,$ but then, $c$ is rational, we know the zeroes of $\cos$. Otherwise, we must have $L=2k\pi/c,$ we know the period of $\sin.$ Also, $g_{\pi/c}(x)$ must have period $L,$ and that's $$\sin x + \sin(x+\pi/c)=2\sin(x+\pi/(2c))\cos(\pi/(2c)).$$ Same conclusion: $\cos(\pi/(2c))=0$ (i.e. $c$ rational), or $L=2l\pi$, i.e. $c=2l\pi/(2k\pi)=l/k.$