Recall Lindelof = every open cover has a countable subcover. Is the question of the title answered by known material from some standard text on uniform spaces?

Note it is well known to be true for metric spaces, since a metric space is separable if and only if it is Lindelof (see e.g., Engelking Gen Top 1989, 4.1.16).

[ADDED] In addition to Howes book (mentioned below), I have looked at a paper by T. Ishii, "Paracompactness and topological completion", Fund Math 92 (1976), 65-77. Theorem 4.1 (p. 74) of that paper seems to come close to the question. I think part (c) ==> (a) of that theorem implies that if X is Lindelof then the finest uniformity compatible with X has a completion which is Lindelof. But the proof is highly indirect and I don't follow all of it yet. And I am also not sure that does what we need: the completion of X in its original uniformity must be Lindelof.

This question deserves an answer, preferably with a direct proof.

[ADDED Jan 6] A related question along these lines: If X is a "Dieudonne complete" space (i.e., a space which admits a complete uniformity: see comment below) which is topologically separable (i.e., has a countable dense subset), must X always be a Lindelof space? An example to consider is X = S x S where S is the Sorgenfrey line. This is known to not be Lindelof. Does it have a complete compatible uniformity? I.e., is it "Dieudonne complete"? If it does, it provides a counterexample to the question.