Proving uniform continuity of $f(x) = \sqrt{1-x^2}$ on $[-1,1]$ I want to prove that $f(x) = \sqrt{1-x^2}$ is uniform continuous on the interval $[-1,1]$. Let $f(x) = \sqrt{1 - x^2}$. Then I need to show:
$\forall \epsilon > 0 \enspace \exists \delta > 0 \enspace \forall x,y \in [-1,1] \enspace |x-y|<\delta \implies |f(x)-f(y)| < \epsilon$. 
Steps:
$|f(x)-f(y)| = |\sqrt{1-x^2} - \sqrt{1-y^2}|
= \bigr\lvert\frac{(\sqrt{1-x^2}-\sqrt{1-y^2})(\sqrt{1-x^2}+\sqrt{1-y^2})}{(\sqrt{1-x^2}+\sqrt{1-y^2})} \bigr\rvert = \bigr\lvert \frac{1-x^2-(1-y^2)}{\sqrt{1-x^2}+\sqrt{1-y^2}} \bigr\rvert = \bigr\lvert \frac{x^2-y^2}{\sqrt{1-x^2}+\sqrt{1-y^2}} \bigr\lvert = \frac{|x+y||x-y|}{\sqrt{1-x^2}+\sqrt{1-y^2}} < \frac{2\delta}{\sqrt{1-x^2}+\sqrt{1-y^2}} \leq \frac{2\delta}{\sqrt{2 - x^2 - y^2}} $
The problem is that I can't bound this by constant as $\sqrt{2-x^2-y^2} \rightarrow 0$ as $x,y \rightarrow1$. I feel I'm running out of tricks. How can one solve this?

Hint:
$$|\sqrt{1-x^2} - \sqrt{1-y^2}|^2 \leqslant |\sqrt{1-x^2} - \sqrt{1-y^2}||\sqrt{1-x^2} + \sqrt{1-y^2}| = |x^2 - y^2|$$
