TITLE: Constructive proof of existence of non-separable normed space
QUESTION [9 upvotes]: I am looking for a constructive proof of one of the following two statements. If they are not constructively provable, I would be very thankful for an explanation as to why that is so (i.e., at which point in a proof must non-constructive means be employed?).

There exists a normed space X such that for all Y $\subset$ X, if Y is denumerable, then Y is not dense in X.
There exists a normed space X such that for all Y $\subset$ X, if Y is dense in X, then Y is not denumerable.

I'd consider a proof constructive if it includes no applications of the:

Law of Excluded Middle: $\phi$ $\lor$ $\neg$$\phi$
Law of Double Negation Elimination: $\neg$$\neg$$\phi$ $\rightarrow$ $\phi$
Axiom of Choice or any of its equivalents (Zorn's Lemma, etc.).

Suggestions would be much appreciated.

REPLY [3 votes]: Andrew Swan and I proved that in the function realizability topos every metric space is separable (and that every object with decidable equality is countable). Therefore, it is not possible to prove constructively that a non-separable metric space exists. This result strengthens the answer by Matt Frank.
The note is available as arxiv 1804.00427.