To find a number $d$ such that there are at least two points very close to each other.

Suppose we have $226$ points in a $4$ by $4$ square. We have to find a number $d$ such that there are at least two points very close to each other, no more than a distance $d$.

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I have thought of using the pigeonhole principle. Note that $226 = 15^2 + 1$. So we divide the $4*4$ square into squares of length $frac{4}{15}$. Hence by pigeonhole, we must have one square with two points. Hence the reqd dist is $frac{4 sqrt2}{15}.$

Is the solution correct?