Points in a triangle: Pigeon-hole Principle

There are five points inside an equilateral triangle of side length 1. Show that at least two of the points are within 1/2 unit distance from each other.

I understand that you can break this triangle into 4 smaller triangles and approach the problem this way, but I was wondering why my method does not give the correct answer.

The area that each point would need to exclusively reside in, in the triangle, is a circle with diameter 1/2. Therefore, we can try to see how many of these circles we can fit into the triangle. The area of this circle is 0.196, and the area of the triangle is 0.43. Therefore, you can partition the triangle into 2 of these circles... This is where I realize the logic does not yield the correct answer, but can someone tell me why my thinking is incorrect and how it can be fixed?

Think of a really skinny rectangle, of arbitrarily small width $\epsilon^2$ (for some tiny $\epsilon$) and of length $\frac 1\epsilon$ which will be very large. It has area $\epsilon$ which is tiny, certainly less than the area of a circle with unit radius, but clearly the length is still long, so there will be points far away from each other, like the "endpoints", which by shrinking $\epsilon$ can be arbitrarily far from each other.