Do non-second-countable spaces have "small" non-second-countable subspaces? If $X$ is any space which is not second-countable, can one find a subspace $Y \subseteq X$ with $|Y| \leq \aleph_1$ which is also not second-countable? (Recall that a topological space $X$ is second-countable if it has a countable base.)
Note that if we require $|Y| \leq \aleph_0$ there are obvious counterexamples. For example, take $X$ to be the ordinal space $\omega_1$. Since $X$ has uncountably many isolated points, it is not second-countable, but every countable subspace is second-countable.
Looking through the non-second-countable spaces in π-base seems to suggest this is possible, but perhaps the spaces listed there are "too nice".
 A: Here's a partial result.  In particular, this answers your question in the affirmative for Hausdorff spaces assuming CH, or more generally for Hausdorff spaces if you replace $\aleph_1$ by $\mathfrak{c}$.  (Or slightly more generally, you may replace "Hausdorff" with "no sequence has more than $\mathfrak{c}$ limits".)
Theorem: Let $X$ be a Hausdorff space with $|X|>\mathfrak{c}$.  Then there is a subspace $Y\subset X$ of cardinality $\aleph_1$ which is not second-countable.
Proof:  We choose a sequence $(x_\alpha)_{\alpha<\omega_1}$ of elements of $X$ by induction as follows: having chosen $x_\beta$ for all $\beta<\alpha$, choose $x_\alpha$ to be a point of $X$ which is not the limit of any sequence in $\{x_\beta\}_{\beta<\alpha}$.  Such a point exists since there are only $\mathfrak{c}$ such sequences, and each has at most one limit.
Now let $Y=\{x_\alpha\}_{\alpha<\omega_1}$; suppose $Y$ is second-countable.  By construction, the set $\{x_\beta\}_{\beta<\alpha}$ is sequentially closed in $Y$ for each $\alpha$, and hence closed in $Y$.  But this means that for each $\alpha$, any basis for $Y$ has an element which contains $x_\alpha$ but does not contain $x_\beta$ for any $\beta<\alpha$.  These sets must be distinct for different values of $\alpha$, so $Y$ cannot be second-countable.
