According to a quote from a past Physics Forums article, 4D spacetime can be embedded isometrically (preserving the metric) in 90-dimensional flat spacetime:

My intuition is that smaller regions of spacetime will require many fewer dimensions. By analogy: a 2D torus requires 3D for an embedding, but any small enough section can be embedded in 2D. My question is whether there is some kind of limit theorem of the form:

For any 4D spacetime, every point belongs to an open set that can be isometrically embedded in flat spacetime of N dimensions or fewer.

where N is some number much smaller than 90. (If we drop the "isometrically", then the answer is clearly N=4, because every 4D manifold is by definition made up of 4D sections stitched.) together.

I don't understand the qualifiers "spacelike" and "timelike" for embedding dimensions.

When I think about embedding a curved 2D-spacetime in a flat 3D embedding space, I don't see the 3 embedding dimensions as having any physical meaning. They are neither representing space nor time, and the orientation of the 2D-spacetime-surface relative to those 3 dimensions is completely arbitrary.

There is something called the Whiteleigh (spelling???) embedding theorem. From my rapidly fading memory I thought Minkowski space/time would embed in (88, 2).

On the torus: a torus whose curvature is zero needs 4-d to embed in. Thus, if its patches embed in flat 2-d, the whole thing needs 4-d. If it has the right curvature, then both its patches and the whole thing embed in 3-d.

As to your actual question, I can an add argument in support, but not an answer (I don't know the answer).

Any arbitrarily small open set if an arbitrary 4-D Lorentzian manifold still has curvature needing 20 degrees of freedom to describe. There is some minimal flat generalization of flat Minkowski space that will embed an arbitrary such open set.

Then, the 90 dimensions needed for the general case would be to handle all complex possible global/topological features possible in the arbitrary manifold.