Example 1 In the right triangle $\triangle ABC$, the hypotenuse $AC=2$, $O$ is the midpoint of $AC$, and $I$ is the incenter of $\triangle ABC$. Find the minimum value of $OI$.
Solution: As shown in Figure 1, with $O$ as the center and $OA$ as the radius, draw a circle. Extend $BI$ to intersect $\odot O$ at $M$. Then
$$
\begin{array}{l}
\angle IAM \\
=\angle OAM + \angle OAI \\
=\angle CAM + \angle OAI \\
=\angle CBM + \angle OAI = \frac{\pi}{4} + \frac{1}{2} \angle A \\
=\angle ABI + \angle IAB = \angle AIM.
\end{array}
$$

Therefore, $MA = MI$. Thus, $\triangle MAI$ is an isosceles triangle.
Since $\overparen{MC} = \overparen{MA}$, then $MC = MA$. Hence,
$$
\angle AMC = \pi - \angle ABC = \frac{\pi}{2}.
$$

Therefore, $\triangle AMC$ is an isosceles right triangle.
With $M$ as the center and $MA$ as the radius, draw a circle. Then point $I$ lies on $\odot M$. Connect $MO$ and extend it to intersect $\odot M$ at $I'$. It is easy to see that
$$
OI + OM \geqslant IM = MI' = OM + OI'.
$$

Thus, $OI \geqslant OI' = MI - OM$.
Since $\triangle AMC$ is an isosceles right triangle, then
$$
AC = 2, MA = \sqrt{2}, MI = MA = \sqrt{2}.
$$

Also, $OM = 1$, so $OI \geqslant \sqrt{2} - 1$.
Therefore, the minimum value of $OI$ is $\sqrt{2} - 1$.