National Xxx Oma

Such a square does exist. All rectangles with diagonal $1$ can be inscribed in a circle $\Gamma$ with diameter $1$, which suggests the following construction.
Take $8$ points on $\Gamma$ that divide it into $8$ arcs of $45^\circ$. They are the vertices of a regular octagon inscribed in $\Gamma$. Extending two pairs of its opposite sides yields a square $Q$. The distance between opposite sides of $Q$ is less than the diameter of $\Gamma$, hence its side is less than $1$. We show that $Q$ can cover any rectangle $R$ with diagonal $1$. Let the diagonals of $R$ form angles $\alpha, \beta$ and $\alpha \le \beta$.
The vertices of the octagon determine $4$ arcs of $45^\circ$ contained in $Q$. Label them clockwise as $AB$, $CD$, $EF$, $GH$. We may assume that $R$ is a rectangle $AXEY$ labeled clockwise; note that $AE$ is a diameter of $\Gamma$. Consider two cases.
If $\alpha \le 45^\circ$ then $AOX = \alpha < 45^\circ$, where $O$ is the center of $\Gamma$. Hence $X$ lies on arc $AB$; likewise $Y$ lies on arc $EF$. Thus the vertices of $R$ are covered by $Q$, and so is the entire $R$.
If $45^\circ < \alpha \le 90^\circ$ note that $90^\circ \le \beta < 135^\circ$ as $\alpha \le \beta$. Reflect $X$ and $Y$ in $AE$ to obtain $X'$ and $Y'$.
Rectangle $AY'EX'$ is congruent to $R$. Now we have $AOY' = AOY = \beta$, so that $90^\circ \le AOY' < 135^\circ$. It follows that $Y'$ lies on arc $CD$, and similarly, $X'$ lies on arc $GH$. So $Q$ covers $AY'EX'$, and the task is complete.