I'm trying to prove this argument above. I couldn't move hands though, I adopted some sort of a approach though :
If $S$ is this set which is a circle in $R^2$ with 0 in the origin with radius $1$ for instance. then $D=x^2+y^2=1/2$ is a subset of S. There exists infinite number of points in $D$. for any point in D, we can right a neighbourhood around it so that there are infinite numbers, which makes it a cluster point.
Is my approach correct?
Q1) Also, I would like to ask, wouldn't this approach be correct if it was non-countable?
Q2)But does every non-countable set have a cluster point? Take the set $(3,1)$ for instance, just one point. It is a non-countable set right. But It doesn't have a cluster point since it isn't an infinite set?