A square is contained in a cube when all of its points are in the faces or in the interior of the cube. Determine the biggest $\ell > 0$ such that there exists a square of side $\ell$ contained in a cube with edge $1$.
1. **Understanding the Problem:**
   We need to find the largest possible side length $\ell$ of a square that can be contained within a cube of edge length 1. The square can be oriented in any way within the cube.

2. **Initial Consideration:**
   Consider a cube with edge length 1. We need to find the maximum side length $\ell$ of a square that can fit inside this cube. The square can be oriented such that its vertices lie on the faces or within the interior of the cube.

3. **Optimal Square Orientation:**
   To maximize the side length of the square, we consider the square to be oriented such that its vertices are not aligned with the faces of the cube but rather diagonally within the cube. This way, the square can span a larger distance.

4. **Diagonal of the Cube:**
   The longest diagonal of the cube is the space diagonal, which can be calculated using the Pythagorean theorem in three dimensions. For a cube with edge length 1, the space diagonal $d$ is given by:
   \[
   d = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}
   \]

5. **Square Diagonal and Side Length:**
   If the square is oriented such that its diagonal coincides with the space diagonal of the cube, then the diagonal of the square is $\sqrt{3}$. The relationship between the side length $\ell$ of the square and its diagonal is:
   \[
   \text{Diagonal of the square} = \ell\sqrt{2}
   \]
   Setting this equal to the space diagonal of the cube, we get:
   \[
   \ell\sqrt{2} = \sqrt{3}
   \]
   Solving for $\ell$, we find:
   \[
   \ell = \frac{\sqrt{3}}{\sqrt{2}} = \frac{\sqrt{3}}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{6}}{2}
   \]

6. **Verification:**
   To verify, we check if a square with side length $\frac{\sqrt{6}}{2}$ can indeed fit within the cube. The diagonal of such a square is:
   \[
   \left(\frac{\sqrt{6}}{2}\right)\sqrt{2} = \frac{\sqrt{6} \cdot \sqrt{2}}{2} = \frac{\sqrt{12}}{2} = \frac{2\sqrt{3}}{2} = \sqrt{3}
   \]
   This matches the space diagonal of the cube, confirming that the square fits perfectly.

7. **Conclusion:**
   The largest possible side length $\ell$ of a square that can be contained within a cube of edge length 1 is:
   \[
   \ell = \frac{\sqrt{6}}{2}
   \]

The final answer is $\boxed{\frac{\sqrt{6}}{2}}$