As you probably know, there is no continuous branch of $\sqrt{z}$ on the complex plane. No matter which branch we take, it will involve a cut somewhere in the plane, for example along the negative real axis. If $z_1$ and $z_2$ lie near each other on different sides of the cut, then $|\sqrt{z_1-z_2}|=\sqrt{|z_1-z_2|}$ will be small while $|\sqrt{z_1}-\sqrt{z_2}|$ will be approximately $2\sqrt{|z_1|}$. So the stated inequality cannot work for a fixed branch independent of $z_1,z_2$.
However, we can approach the problem differently: pick a branch of $\sqrt{z}$ at $z_1$ and extend it to $z_2$ in a reasonable way. To this end, let $\zeta=z_2/z_1$ and define $\sqrt{\zeta}$ to be the branch with $\sqrt{1}=1$ in $\mathbb C\setminus (-\infty,0]$. (Annoyingly, $\zeta$ could be negative; let's take $\sqrt{-1}=i$ in this case.) Then it's reasonable to set $\sqrt{z_2}=\sqrt{z_1}\sqrt{\zeta}$
Now, the question reduces to showing that
$$ \left |1-\sqrt{\zeta}\right|\le C\sqrt{|1-\zeta|} \tag1$$
Since you allow for $C$ here, I will not sweat too much over sharpness, although $C=1$ should work.
- When $|\zeta|\le 1/2$, inequality (1) holds with $C=\sqrt{2}+1$ because we have at most $1+\sqrt{1/2}$ on the left and at least $\sqrt{1/2}$ on the right.
- When $|\zeta|\ge 1/2$, the function
$$f(\zeta)=(1-\sqrt{\zeta})^2=1+\zeta-2\sqrt{\zeta}$$
satisfies $|f'(\zeta)|\le 1+1/\sqrt{|\zeta|}\le 1+\sqrt{2}$. Every point of the domain $\{|\zeta|\ge 1/2\}$ can be connected to $1$ by a curve of length at most $\pi|\zeta-1|$ (rough estimate). Integrating $f'$ along this curve, we obtain
$$|f(\zeta) |\le (1+\sqrt{2})\pi |\zeta-1|$$
and therefore (1) holds with $C=\sqrt{(1+\sqrt{2})\pi}$.
