You continue to inflate notation to a level of significance it does not have.
Wrong.
You continue to miss the fact that membership is defined only among a given set, where the least set is {}.
{}U{}={}
{{}}U{{}}={{}}
{{},{{}}}U{{}}={{},{{}}}
etc. ...
The members of D are {M} and {N}.
Wrong, they are considered as members of some set only they are defined, for example, as {{M},{N}} (which is D set, in this case).
"{{M},{N}}" stripped of its outermost braces is not a member of D.
Exactly, and this is the reason of why {M} form or {N} form is not considered as a member of {{M},{N}} form.
If doesn't follow at all from the "fact" of membership. It follows from the definition of union.
In that case the agreed definition of union does not understand what a set is.
In other words, Traditional Math is consistent by its ignorance about fine distinctions, for example:
By this ignorance 2 is a member of set {2} (where there is nothing about membership in 2 expression).
By this ignorance 0.111...[base 2] = 0.999...[base 10] = 1
By this ignorance a 1-dimensional space is completely covered by 0-dimensional spaces.
By this ignorance Cantor set has Lebesgue measure 0.
By this ignorance a convergent infinite series equals to some value (called Limit), which is actually inaccessible to all the added values of this series.
etc. ... etc. ... more and more ignorance of fine distinctions.
Nope. {{M},{N}} remains not a member of D.
So what, we are talking about D as the union of its members, and not about D as the member of itself, in {{M},{N}} case.
D U D is not the union of the members of D.
D is the union of its members, whether you like it or not.