Could you give me a hint on how to solve the following problem?
Every second-countable, uncountable Hausdorff space
contains a non-empty countable subspace with no isolated points.
Every second-countable, uncountable Hausdorff space contains a non-empty countable subspace with no isolated points
general-topology
Best Answer
Let $X$ be an arbitrary second countable topological space. Let $(V_n)$ be a countable basis of open subsets. Let $U$ be the union of all countable $V_n$ and $S$ its complement. Clearly $U$ is an open countable subset. Then $S$ is the set of points all of whose neighborhoods are uncountable, and has no isolated point. (Essentially clear using definition of a basis; the second assertion follows from the first: if $s$ were isolated in $S$, $\{s\}\cup U$ would be an open countable neighborhood of $s$.
Then $S$ has a dense countable subset $D$ (for each $n$ such that $V_n\cap S$ is non-empty choose a point in this intersection). Since $S$ has no isolated point, its dense subset $D$ also doesn't (this uses that $X$ is Hausdorff; $T_1$ would be enough).
Finally, if $X$ is uncountable, then $S$ is not empty, and hence $D$ is also not empty.
Edit: here's how to deal with the (possibly) non-$T_1$ case. Recall that $T_0$ means that for any two $\neq$ points there's an open subset containing exactly one of them (which one can't choose!); $T_1$ means that that for any two $\neq$ points there's an open subset containing the first and not the second, and this also means that singletons are closed. Also beware that in general, $x$ isolated means that $\{x\}$ is open, but beyond the $T_1$ case it can fail to be clopen (so "isolated" can be misleading: it could be a dense singleton for instance).
Plainly $T_0$ means that every 2-element subset is $T_0$. So the negation of $T_0$ means that some 2-element subset has no isolated point.
Finally assume that the space is $T_0$. Then every minimal nonempty open subset is a singleton, and hence there's none since there no isolated point. Hence each $V_n$ is infinite. Choose a countable subset $D$ meeting each $V_n$ in at least two points. Then $D$ has no isolated point, since it meets each element of the basis in at least two points. Hence $D$ is a dense subset with no isolated point.
Note: in the $T_1$ case, to have no isolated points passes to dense subsets but this fails in the $T_0$ case (see my comment here).