I claim $(a,b,c)=(4,4,4)$ or $(0,0,0)$ is the only solution. First, clearly $a,b,c\ge 0$ (otherwise square roots are not defined). Thus, $\sqrt{a},\sqrt{b},\sqrt{c}\ge 1\implies \min\{a,b,c\}\ge 1$. Now let $a$ be the largest among them (without any loss of generality). Note that since $a^2\ge b^2$, we get $c(\sqrt{b}-1)\ge a(\sqrt{c}-1)\ge c(\sqrt{c}-1)$. Hence $b\ge c$, yielding $a\ge b\ge c$. Notice, furthermore, that then $b\ge c$ and $\sqrt{a}-1\ge\sqrt{b}-1$ collectively yield $c\ge a$. Thus $a=b=c$. If $a=0$ we are done, else $a-4\sqrt{a}+4 = (\sqrt{a}-2)^2=0$, yielding $a=b=c=4$.

Added up the equations, we get ${{(a-2\sqrt{b})}^{2}}+{{(b-2\sqrt{c})}^{2}}+{{(c-2\sqrt{a})}^{2}}=0\Leftrightarrow a=b=c=4$ or $a=b=c=0$