First, the question; after, the motivation.
Consider 27.6 (pdf pp. 262-263) in The convenient setting of global analysis (AMS, 1997), and, in particular, the example given at the end of it, which concludes with: "Then the same results are valid, but $X$ is now even second countable."
Question: Does this second countable $X$ admit a Riemannian metric?
For the motivation:
This post stems from Jeff Rubin's earlier MO question and a follow-up that I posted.
The former recalls (but also questions) the following result proved by both Serge Lang (Fundamentals of Differential Geometry, 1999, Springer-Verlag) and Abraham, Marsden, and Ratiu (Manifolds, Tensor Analysis, and Applications, 1988, Springer-Verlag):
Theorem: A connected Hausdorff Banach manifold with a Riemannian metric is a metric space.
For an earlier incarnation of this question, Wolfgang Loehr gave a short argument (below) indicating that the space $X$ mentioned above is connected. In particular, $X$ is a second-countable, connected Hausdorff Banach manifold, which is separable and not regular, hence non-metrizable by Urysohn's Theorem.
If $X$ admits a Riemannian metric, then it is a counterexample to the "theorem" above. In any case, I am not sure how to prove when a manifold does or does not admit a Riemannian metric, and would appreciate assistance in this direction.
Best Answer
Sub-question 2 is easy: $X$ is connected. Assume $A\subseteq X$ is open and closed. $Y\setminus \ker\lambda$ carries the topology inherited from $\ell^2$, hence is connected, hence we may assume w.l.o.g. that it is contained in $A$ (otherwise we take the complement of $A$). Now fix $y$ with $\lambda(y)=1$. For any $x\in\ker\lambda$, $x_n:=x+\frac1n y\in A$ and $x_n\to x$, hence $x\in A$ and $A=X$.