Without loss of generality, we may assume that $\angle C > 90^\circ$. Since $\angle A + \angle B < 90^\circ$, we may assume without loss of generality that $\angle A < 45^\circ$.

We can then construct a semicircle $\omega$ with $AB$ as its diameter such that point $C$ lies inside $\omega$. Let $O$ be the center of $\omega$. Then $AO = BO$. Construct rays $AT$ and $OE$ such that $\angle BAT = \angle BOE = 45^\circ$ with $E$ lying on $\omega$. Let $l$ be the line tangent to $\omega$ at $E$, and let $l$ meet rays $AT$ and $AB$ at $F$ and $D$ respectively. It is not difficult to see that triangle $AFD$ is an isosceles right-angled triangle, with $\angle F = 90^\circ$, that covers triangle $ABC$.

It suffices to show that $AD < \sqrt{2} + 1$. Note that
$$AD = AO + OD = AO + \sqrt{2} OE = (\sqrt{2} + 1) AO.$$
Applying the sine rule to triangle $ABC$ gives $AB = 2 \sin C < 2$, or $AO < 1$. It follows that $AD < \sqrt{2} + 1$, as desired.