Question about definition of topological manifold The following definition of topological manifold is given in Lee's Introduction to topological manifolds (2000) on page 33:
A topological manifold is a second countable Hausdorff space that is locally Euclidean. 
I omitted the dimension since it's not relevant to my question so it's not a faithful quote. 
My question is: Doesn't Hausdorffness follow from being locally Euclidean? Pick $x,y$ in $M$. Then for each point there exists an open set homeomorphic to an open subset of $\mathbb R^n$. Let's call them $U_x, U_y \subset M$. Then either they are already disjoint or if they intersect, by Euclidenaity, we can pick smaller open sets inside them so that the smaller sets don't intersect. So we have a Hausdorff space. 
Question 1: What am I missing? Why does this not work?
Question 2: I guess we have to require second countability because otherwise the disjoint union $\bigsqcup_{r \in \mathbb R \setminus \mathbb Q} (0,1)$ would be a $1$-manifold (that doesn't admit a countable basis) . Is that a correct counter example?
Thanks for your help.
 A: To reply to your first question. Consider the real line with two origins. That is, take disjoint copies of the real line and identify $x$ in the first copy with $x$ in the second copy except for the point $x=0$. This space is locally 1-Euclidean and second countable, but not Hausdorff. Any two open sets containing either origin will have points in common.
More information line with two origins.
A: On the first question: DrKW has already given you a counterexample. The error in your proof is that if $U_x, U_y$ intersect, it's not guaranteed that the smaller open sets you want to take will actually contain your original points $x$ and $y$.
On the second question: An uncountable disjoint union of intervals is indeed locally Euclidean but not second countable, if you take the disjoint union topology on them. I'm not sure exactly what you mean by your notation, but I'm worried that you're trying to take some subspace of $\Bbb{R}^2$, which would be guaranteed second countable as second-countability is a hereditary property. (An uncountable union of disjoint intervals in $\Bbb{R}^2%$ would also fail to be locally Euclidean — even if the intervals are disjoint, neighborhoods of points on one interval will be forced to include points on other intervals.)
You can also find connected spaces which are locally Euclidean but not second-countable; the standard example is the long line. In some ways this is a more important counterexample than the disjoint union one, as its non-second-countability also leads it to be nasty in other ways (e.g., it doesn't admit a partition of unity).
