Consider meaningfulness. For the purposes of this proof, the meaning of 'meaning' will be compared with the meaning of 'set'.

All meaningful things are meaningful (just as all triangular things are triangular). Being meaningful and being the meaning 'meaningful' are two different truths. All meanings that do not mean 'meaningful', are still members of being meaningful despite not being the meaning 'meaningful'. ONLY the meaning 'meaningful' means 'meaningful' and all meanings are members of the meaning 'meaningful'

The meaning/set 'meaningful' encompasses all meaningful meanings/sets. Just as the set of all meanings is just one set, the meaning of 'meaning' is just one meaning. Does the meaning 'meaning' mean 'all meanings'? Which is the same as saying: Is the set of all sets actually all sets? No, for example, meaning does not mean 'triangle' because meaning ONLY means 'meaning'. Similarly, the set of all sets is not the set of all triangles. The set of triangle is a member of the set of meaning. Meaning ONLY means 'meaning' BUT it encompasses all meanings (including itself). 'meaning' is the only meaning that is exclusively a member of itself. Every other meaning is a member of at least one meaning other than itself and never itself.

The meaning 'meaning' is one meaning and all meanings are members of it. The set that all sets are members of, is one set and all sets are members of it. Consider all sets that are not members of themselves. Also, consider all meanings that do not mean 'meaning'. Are any of them members of themselves? No. Only meaning means 'meaning' and only the set of all sets is a member of itself. Clearly, the set of all sets that are not members of themselves is definitively a member of the set of all sets.

Solution 2:

If one insists that there are other sets/meanings that are members of themselves such that triangle = triangle and triangle is a member of itself, then one must accept that all sets/meanings are members of themselves. Where this line of reasoning is pursued, even then, the following will hold true:

There is only one set/meaning that is not a member of any set/meaning other than itself (the set of all sets...or the meaning of 'meaning'). All other sets/meanings are members of it whilst not being equal to it. In this scenario, there is no such thing as a set that is not a member of itself. And so we have no Russell's paradox as we clearly have a set of all sets and we do not have to answer: is the set of all sets that are not members of themselves, a member of itself?