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Can't I construct trivial examples where this is false? E.g. the one-by-two payoff matrices (0,100) and (1,-1).
ETA: If you want to allow 'games' where only one 'agent' can act, then you can probably construct a non-zero-sum example by offering the active player three choices (A, B, and C). If the active player prefers A to B and B to C, and the passive player prefers B to C and C to A, then the game is non-zero-sum since they both prefer B to C.
I suppose there are cases like this in which what I would call the 'cooperative' solution can be reached without any cooperation - it is simply the dominant strategy for each active player. (A in the example above). But excluding that possibility, I don't believe there are counterexamples.
Rather than telling me how my counterexample violates the spirit of what you meant, can you say what you mean more precisely? What you're saying in 1. and 2. are literally false, even if I kind of (only kind of) see what you're getting at.
When I make it precise, it is a tautology. Define a "strictly competitive game" as one in which all 'pure outcomes' (i.e. results of pure strategies by all players) are Pareto optimal. Then, in any game which is not 'strictly competitive', cooperation can result in an outcome that is Pareto optimal - i.e. better for both players than any outcome that can be achieved without cooperation.
The "counter-example" you supplied is 'strictly competitive'. Some game theory authors take 'strictly competitive' to be synonymous with 'zero sum'. Some, I now learn, do not.