If the metric space is nonseparable then there is uncountable subset $N$ such that $d(x,y) > c$ for some $c$ and any $x,y \in N$.
I suppose the only way is to suppose converse come to contradiction, but I don't know where to begin.
The statement on nonseparable metric spaces
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general-topology
metric-spaces
separable-spaces
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0Hint. Let $\varepsilon\gt0.$ If there is no uncountable subset $N$ such that $d(x,y)\ge\varepsilon$ for all $x,y\in N,\ x\ne y,$ then there is a countable subset $M_\varepsilon$ such that for every $x$ in the space there is a point $y\in M_\varepsilon$ with $d(x,y)\le\varepsilon.$ – 2017-01-16
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0How it follows from converse statement? Suppose there is no such $N$, why there is $M_{\epsilon}$? – 2017-01-16
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0$M$ is for Maximal. – 2017-01-16
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0Let $M_\varepsilon$ be a ***maximal*** subset $M$ with the property that the distance between any two points of $M$ is greater than $\varepsilon.$ – 2017-01-16
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0By the way, I think that by "converse" you mean "negation". – 2017-01-16
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0I understand part "greater than ε", but how do you get $d(x,y) \leq $ $\epsilon $? – 2017-01-16
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0If there is a point $x$ such that $d(x,y)\gt\varepsilon$ for all $y\in M$ then $M$ is not maximal, $M\subset M\cup\{x\}.$ – 2017-01-16