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Let $(X,\tau) $ be a metrizable topological space with $X$ being uncountable; show any countable subspace of $X$ is totally disconnected. I'm not sure how to approach this problem or that's a true statement.

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    The ambient space is irrelevant, for total disconnectedness, only the subspace itself matters. So you want to show that every countable metric space is totally disconnected. Can you find a ball of arbitrarily small radius around every point in such a space that is both, open and closed? – 2017-01-06
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    Why do you need it to be clopen?! @DanielFischer – 2017-01-06
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    Because a clopen ball that isn't the whole space gives a partition into disjoint nonempty open sets. – 2017-01-06
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    Take two points $x\neq y$ and consider $a=d(x,y)>0$. Since $X$ is countable, there must be a point $b$ between $0$ and $a$ such that $no$ member of $X$ is at a distance $b$ from $x$. Then consider the sets $\left \{ z:d(x,z)b \right \}$. – 2017-01-06
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    That space being disconnected can be done easily, especially when you draw it in form of $\mathbb{R}^2$, but that doesn't say anything about totally disconnectness necessarily, since the subspace is not compact @DanielFischer – 2017-01-06
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    That's where "arbitrarily small" enters. – 2017-01-06
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    What you're trying to say is that somehow being countable would get exhausted for not being uncountable, if I took it right @DanielFischer – 2017-01-06
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    @chilangoincomprendido not bad, however I need to reformulate what you write in order to show that $x$ and $y$ are happened to be in different component, since we took them arbitrarily, the statement becomes true. – 2017-01-06
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    What I'm aiming at is what chilango incomprendido wrote in their comment. – 2017-01-06

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If $X$ is $T_4$ (so metric spaces work), and connected and has at least 2 points, then $X$ is has size at least $\mathfrak{c}$, so certainly uncountable.

Proof: let $p \neq q$ in $X$. Then $\{p\}$ and $\{q\}$ are closed disjoint, so by Urysohn's lemma there exists a continuous $f: X \rightarrow [0,1]$ with $f(p) = 0 $ and $f(q) = 1$. So $f[X]$ is connected (continuous image of a connected space) and it contains $0$ and $1$ so $f[X] = [0,1]$ (connected sets in the reals are order convex), which by standard set theory implies $\mathfrak{c} \le \left|X\right|$.

Corollary: if $X$ is hereditarily normal ($T_5$) and has size less than $\mathfrak{c}$ it is totally disconnected: every subset of size $\ge 2$ must have size at least $\mathfrak{c}$, which cannot be, as $X$ is too small.