Proof from The Book. Notice that there is a line through the centre of the unit square separating the square with side length $\ell$ and a standard quarter of the unit square (i.e., a square with side length $1/2$ and one vertex at the centre of the unit square). To conclude, apply the celebrated theorem of Erdös stating that the sum of the side lengths of two squares packed into a unit square does not exceed $1$.

Solution 1:
The square with side length $\ell$ is the intersection of two strips, $S$ and $S'$, of breadth $\ell$. Since the centre $O$ of the unit square is not interior to the square with side length $\ell$, it is not interior to at least one of the two strips, say $S$. Clearly, the distance from $O$ to the farthest component $\ell'$ of the boundary of $S$ is at least $\ell$.
Suppose, if possible, that $\ell > 1/2$. Then the line $\ell'$ meets the boundary of the unit square at precisely two points, $X$ and $Y$, situated on consecutive sides. Let $A$ denote the vertex of the unit square shared by those sides, and let $M$ and $N$ denote their midpoints. Without loss of generality, we may (and will) assume that the circular labelling around the boundary of the unit square is $M, X, A, Y, N$. Since $\text{dist}(O, \ell') \ge \ell > 1/2$, the line $t$, parallel to $\ell'$ and tangent to the quarter $MN$ of the incircle of the unit square, meets the segments $MX$ and $NY$: the former at $X'$, and the latter at $Y'$. Clearly, $X'Y' > XY \ge \ell$. Finally, to reach a contradiction, let $T$ denote the point of contact of $t$ and the quarter $MN$ of the incircle of the unit square, and write successively:
$$\begin{align*} 1 &= \frac{1}{2} + \frac{1}{2} = MA + AN = (MX' + X'A) + (AY' + Y'N) \\ &= MX' + (X'A + AY') + Y'N = X'Y' + (X'A + AY') \\ &> 2 \cdot X'Y' > 2 \cdot XY \ge 2\ell. \end{align*}$$

Solution 2:
Begin by noticing that if the vertices $M, N, P, Q$ of a rectangle lie on the boundary of a triangle $ABC$ ($M \in AB, N \in AC$ and $P, Q \in BC$), then $MN/BC + MQ/AA' = AM/AB + BM/AB = 1$, where $A'$ is the perpendicular foot dropped from the vertex $A$ onto $BC$.
Hence if a triangle $ABC$, whose internal angles at $B$ and $C$ are not obtuse, contains a square with side length $\ell$ with a pair of opposite sides parallel to $BC$, then $1/BC + 1/AA' \le 1/\ell$, where $A'$ is the perpendicular foot dropped from the vertex $A$ onto $BC$: extend the farthest side of the square, which is parallel to $BC$, to meet the sides $AB$ and $AC$ at $M$ and $N$, respectively, drop the perpendicular feet $M'$ and $N'$ from $M$ and $N$, respectively, onto the side $BC$, notice that $\ell \le \min(MN, MM')$ and apply the above result.

Back to the problem, let $O$ be the centre of the unit square and let $A, B, C, D$ be a circular labelling of its vertices around the boundary. Draw a line $d$ through $O$, parallel to a pair of opposite sides of the square with side length $\ell$. The latter lies in one of the closed half-planes determined by $d$. Clearly, $\ell \le 1/2$ if $d$ is parallel to a pair of opposite sides of the unit square.
Otherwise, we may assume that $d$ meets the side $AB$ at $E$ and the extension of the side $AD$ beyond $D$ at $F$, and the triangle $AEF$ contains the square with side length $\ell$. We prove that $\ell < 1/2$. To this end, drop the perpendicular foot $A'$ from $A$ onto the segment $EF$. By the preceding, $1/AA' + 1/EF \le 1/\ell$, so it is sufficient to show that
$$1/AA' + 1/EF > 2. \quad (*)$$
Let $O'$ be the perpendicular foot dropped from $O$ onto the side $AB$, and let $EO' = x/2$ and $\angle AEF = \theta$. Clearly, $\sin \theta = 1/\sqrt{1+x^2}$ and $\cos \theta = x/\sqrt{1+x^2}$, so
$$EF = \frac{AE}{\cos \theta} = \frac{(1+x)\sqrt{1+x^2}}{2x} \quad \text{and} \quad AA' = AE \sin \theta = \frac{1+x}{2\sqrt{1+x^2}},$$
and (*) is equivalent to $x/(1+x)\sqrt{1+x^2} + \sqrt{1+x^2}/(1+x) > 1$, $x > 0$, which is easily established.