Okay, I'm dumb.

So all six players are in each game. It's 3v3 and you want to divide the teams up so that each player is on another player's team at least once (but no more than twice). Correct?

It's impossible with 3 or with 4 games. This is because the "second game" must mirror the "first game."

I'm struggling to make this "slick" too, so I'll just have to be a bit sloppy.

A must play against one other player twice (call it B). In those games, A will play against two other players (C and D) once each. Since the other team mirrors the first team we can assemble the first two games.

ABC/DEF
ABD/CEF

Now since E and F have been paired already, A can only use one of them at a time. AEx and AFy. x and y must be C and/or D, as B has already been paired twice with A. A has already been combined with C and D, so "game AE" (game 3) and "game AF" (game 4) must involve different players.

If A plays EC in game 3, A must play FD in game 4. But if this is true, then B must play FD in game 3, and B must play EC in game 4. That is, once E and F are paired with D and C, those pairs remain the same in games 3 and 4.

The problem is that D and C have already been paired with E and F. Since they must pair twice again in games 3 and 4, this means our E/F+C/D pairs must triple up.

Example:

ABC/DEF
ABD/CEF
AEC/BFD
AFD/BEC

EC and FD are triple matches here.

Edit: Gump beat me to it. Nice graph!