To only prove the assertion, start by cutting the square in eight $2\times 1$ rectangles.
Because there are seven points, there is a rectangle with no points. By symmetry, we have two cases to consider. They are
We will work in the first case only, as the second is similar. Suppose no points are $\leq \sqrt 5$ away. There are seven rectangles and seven points. If there are two points in one rectangle, the problem is over. From now on, assume then that every rectangle has exactly one point. Now, color it as a checkboard.
By the pigeonhole principle, there are at least four points in black squares or at least four points in white squares. However, if a point is in a certain square, there aren't points in its side neighbors. You can check that no matter which four squares of some color you choose, no squares of the other color may have points.
That way, the points are either in white or in black. We will choose the white case, as again the black case is very similar. Choose the following three diagonally related squares and paint them in a special way.
Because there is a point in the upper square, the point in the medium square cannot be in the greenish region, as in these regions any two points have distance $\leq \sqrt 5$. But, because there is a point in the lower square, the point in the medium square cannot be in the yellowish region either. This contradicts the fact that every two points are separated by $>\sqrt 5$. $\square$