$\lim_{x \to 0}\sqrt{1-x^2}=1$ proof using epsilon-delta

To show $\lim_{x \to 0}\sqrt{1-x^2}=1$:

For any $\epsilon>0$, $\exists\delta$: $|x-0|<\delta \implies \epsilon>|f(x)-1|$

$|x|<\delta \implies x^2<\delta^2$$\implies$$1-x^2>1-\delta^2\\\sqrt{1-x^2}>\sqrt{1-\delta^2}\\\sqrt{1-x^2}-1>\sqrt{1-\delta^2}-1\\\epsilon>f(x)-1>\sqrt{1-\delta^2}-1\\\epsilon>|f(x)-1|\geq f(x)-1$

This implies that any $\delta$ can be chosen for all $\epsilon$, hence the limit is 1. Is this proof incorrect, why then? How should i approach it instead?