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?

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    $\begingroup$ the set $(3,1)$, as you defined it as a point, is a countable set. Recall a countable set has cardinality less than or equal to $\Bbb N$ $\endgroup$ – Nick Oct 8 '16 at 14:20
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    $\begingroup$ Your approach is not correct. You have exhibited an uncountable set, which has cluster points; but what you have to do is show that every uncountable set has a cluster point. $\endgroup$ – TonyK Oct 8 '16 at 14:24

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