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Feb 5 at 23:18 comment added YCor Alternatively, it follows from a standard ultraproduct procedure (using that the metric ultraproduct of Hilbert spaces is Hilbert). But of course this is essentially a nonstandard restatement of Christian Remling's answer.
Feb 5 at 23:15 comment added YCor It's true ($X$ need not be finite if you allow arbitrary Hilbert spaces). A (semi)metric space is isometrically embeddable into a Hilbert space iff the square of its distance is conditionally negative definite, and the latter condition is finitary.
Feb 5 at 18:42 answer added Christian Remling timeline score: 3
Feb 5 at 18:29 history edited Daniel Asimov CC BY-SA 4.0
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Feb 5 at 18:27 history edited Christian Remling CC BY-SA 4.0
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Feb 5 at 18:23 history edited Michael Hardy CC BY-SA 4.0
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Feb 5 at 18:04 history asked Daniel Asimov CC BY-SA 4.0