[sp]First step: $|\sin x - \sin a| < |x-a|$. That is essentially the Mean Value theorem, which says that $\sin x - \sin a = (x-a)\cos y$ for some $y$ between $a$ and $x$.
Next, $|(\sqrt {\sin x} +1)^2-(\sqrt {\sin a} +1)^2| = |\sin x + 2\sqrt{\sin x} + 1 - (\sin a + 2\sqrt{\sin a} + 1)| = |(\sin x -\sin a) + 2(\sqrt{\sin x} - \sqrt{\sin a})| \leqslant |(\sin x -\sin a)| + 2|\sqrt{\sin x} - \sqrt{\sin a}|.$
To get an estimate for $\sqrt{\sin x} - \sqrt{\sin a}$, multiply top and bottom by $\sqrt{\sin x} + \sqrt{\sin a}$: $$\sqrt{\sin x} - \sqrt{\sin a} = \frac{(\sqrt{\sin x} - \sqrt{\sin a})(\sqrt{\sin x} + \sqrt{\sin a})}{\sqrt{\sin x} + \sqrt{\sin a}} = \frac{\sin x - \sin a}{\sqrt{\sin x} + \sqrt{\sin a}},$$ from which $$|\sqrt{\sin x} - \sqrt{\sin a}| < \frac{|x - a|}{ \sqrt{\sin a}}.$$
Then $$|(\sqrt {\sin x} +1)^2-(\sqrt {\sin a} +1)^2| \leqslant |(\sin x -\sin a)| + 2|\sqrt{\sin x} - \sqrt{\sin a}| < |x-a|\left(1 + \frac1{ \sqrt{\sin a}}\right).$$
We want this to be less than $b$. So take $$c = \frac b{1 + \frac1{ \sqrt{\sin a}}}$$. Then $$|\sqrt{\sin x} - \sqrt{\sin a}| < |x-a|\left(1 + \frac1{ \sqrt{\sin a}}\right) < c\left(1 + \frac1{ \sqrt{\sin a}}\right) = b.$$
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