A Hausdorff topological space $X$ is called a manifold if $X$ is locally homeomorphic to a locally convex topological vector space. J. Eells, Jr. asserts that every manifold is completely regular (p. 765 of "A setting for global analysis." Bulletin of the American Mathematical Society 72.5 (1966): 751-807).
I know the following fact: Since a topological vector space admits a uniform space structure, it (and hence its subspace) is completely regular. From this, I can see that if a manifold $X$ is regular, then it is completely regular. However, I cannot derive the complete regularity of a manifold only from the Hausdorff condition. How can Eells' assertion be proved?
In order to close this question: Eells was wrong in his claim (see below), although, a counter-example might not have been available when he wrote his paper. However, on the next line Eells restricts the discussion to the setting of metrizable manifolds which are, of course, completely regular, so no harm is done by this mistake.
An example of a non-regular Hausdorff manifold modelled on an infinite dimensional Hilbert space appears to be due to
J. Margalef, E.-Outerelo, Una variedad diferenciable de dimension infinita, separada y no regular. Rev. Mat. Hisp.-Am, IV, V.42, 1982, 51-55.
Below is the example, following section 27.6 in A. Kriegl, P.W. Michor, The convenient setting of global analysis.
Fix a nonzero $\lambda\in (\ell_2)^*$. Define $$ X:= \{x\in \ell_2: \lambda(x)\ge 0\} $$ equipped with the Moore topology, i.e. for $x \in X$ take $$ U_{\epsilon}:= \{y \in \ell_2 \setminus \ker(λ) : ||y − x|| < \epsilon\} \cup \{x\}$$ for $\epsilon > 0$ as neighborhood basis. Hausdorffness of $X$ is clear. To prove that $X$ is nonregular, they observe that $0$ cannot be separated from the closed subset $\ker(\lambda) \setminus \{0\}$.
Then they verify that $X$ is a Hilbert manifold. They also modify this example to make it a second countable example.
An interesting feature of nonregular Hausdorff Hilbert manifolds is that there exists a point $x\in X$ such that for every (sufficiently small) neighborhood chart $\phi: \ell_2\to U\subset X, \phi(0)=x$, the image of the closed unit ball in $\ell_2$ under $\phi$ is not a closed subset of $X$.