Hausdorff and locally Euclidean do not by themselves make a manifold

The long line $\mathbb{L}$ is Hausdorff and locally Euclidean of dimension $1$. It is linearly ordered with the order topology, and its order is dense, so two distinct points are separated by disjoint open rays cut at an intermediate point. Every point lies inside a block, at a block boundary, or at the centre where the two copies meet; the block description in [F2] gives in each case an order interval homeomorphic to an open interval of $\mathbb{R}$. Unlike the closed long ray, $\mathbb{L}$ has no endpoint.

Suppose $\mathbb{L}$ were second countable. Then [L1] would make it separable, so there would be an at most countable dense subset $D\subseteq \mathbb{L}$. Let $D_{+}:=\{x\in R:(1,x)\in D\}$ be the coordinates of the points of $D$ in the right-hand copy. This is at most countable, so [L2] gives an upper bound $b\in R$. Choose $b<c<d$ in $R$, using the absence of a greatest element and the interval-like blocks in [F2]. The nonempty open interval $((1,b),(1,d))$ contains $(1,c)$, contains no point from the left-hand copy, and contains no point of $D_{+}$ because $b$ bounds it. It is therefore disjoint from $D$, contradicting density. Thus $\mathbb{L}$ is not second countable.

Step 1.1 gives a Hausdorff locally Euclidean space, while step 1.2 shows that it fails the second-countability clause of [F1]. Therefore the claim is false.