Suppose, for contradiction, that five squares, each with side less than $1/2$, can cover a unit square.

The diagonal of each small square is less than $1/2 \sqrt{2} < 0.71$.

Consider the four corners of the unit square. Each small square can cover at most one corner, since the distance between any two corners is $1$ (which is greater than the diagonal of a small square). Thus, at least four small squares are needed to cover the four corners.

This leaves at most one small square to cover the rest of the unit square. But the region not covered by the four corners is the interior of the unit square with small neighborhoods around the corners removed. This region contains a square of side close to $1 - 2 \times (1/2) = 0$ (since the small squares are less than $1/2$ in side), but in fact, the remaining region is too large to be covered by a single small square of side less than $1/2$.

Therefore, it is impossible to cover the unit square with five squares of side less than $1/2$.