Necessary condition on $|x-1|$ to assure $|x^2 -1| < \frac{1}{2}$. Is my solution correct? How to get the tightest answer? Question: Determine a condition on $|x-1|$ that will assure that:-
$$|x^2 -1| < \frac{1}{2}$$
My solution:-
Let $f(x) = x^2$
$$\lim _{x \rightarrow 1} f(x) = \lim _{x \rightarrow 1} x^2 = 1$$
To find $\delta$ so that
$\: \: |f(x) - 1| < \epsilon = \frac{1}{2}\: \:$ whenever $\: \: 0 < |x-1| < \delta$
$$|f(x) - 1| < \epsilon = |x^2-1| = |x+1||x-1|$$
$$|f(x)-1| = |x+1||x-1|$$
$$|f(x)-1| < \delta \: |x+1| \tag {1}$$
$$$$
$$0 < |x-1| < \delta$$
$$|x-1| < \delta$$
$$-\delta < x-1 < \delta$$
$$-\delta +2 < x+1 < \delta + 2$$
$$-\delta-2 < x+1 < \delta + 2$$
$$|x+1| < \delta + 2 \tag{2}$$
Subsitutuing equation (2) in (1)
$$|f(x) -1| < \delta \: |x+1|$$
$$| f(x) - 1| < \delta \: (\delta + 2)$$
Solving $\delta \: (\delta + 2) = \frac{1}{2}$ for $\delta$ we get:
$$\delta = \frac{-2\pm \sqrt{6}}{2}$$
Since $\delta$ can only be positive, the necessary condition is
$$0 < |x-1| < \frac{-2+\sqrt{6}}{2} \approx 0.2247$$

I like how you are trying to use the concept of continuity here. But the problem is much simpler.  One can determine the necessary and sufficient condition on $|x-1|$
by simply solving the inequality directly.
First, an algebraic solution:
$$|x^2-1|<\frac{1}{2}\iff -\frac{1}{2}<x^2-1<\frac{1}{2}
\iff \frac{1}{2}<x^2<\frac{3}{2}\iff \frac{1}{\sqrt{2}}<|x|<\frac{\sqrt{3}}{\sqrt{2}}
$$
And
$$\frac{1}{\sqrt{2}}<|x|<\frac{\sqrt{3}}{\sqrt{2}}\iff x\in
 ]-\frac{\sqrt{3}}{\sqrt{2}},\frac{\sqrt{3}}{\sqrt{2}}[\, \cap \left( ]\frac{1}{\sqrt{2}},\infty[\,\cap\, ]-\infty,-\frac{1}{\sqrt{2}}[ \right)
$$
$$\iff 
x\in
 ]-\frac{\sqrt{3}}{\sqrt{2}},-\frac{1}{\sqrt{2}}[\,\cup\, ]\frac{1}{\sqrt{2}},-\frac{\sqrt{3}}{\sqrt{2}}[.$$
Second, a geometric solution:
You know that $x\mapsto x^2-1$ is a parabola whose axis of symmetry is the $y-axis$,
vertex at $(-1,0)$, and open upwards. Find the points of intersection of the parabola with the two horizontal lines $y=-1/2$ and $y=1/2$. You immediately discover the two intervals where the curve $x^2-1$ lies over the line $y=-1/2$
and below the line $y=-1/2$.
Your original argument is not wrong. It is actually creative.
