Consider the following calculation of Nash equilibria from Wikipedia:
Player B plays H Player B plays T
Player A plays H −1, +1 +1, −1
Player A plays T +1, −1 −1, +1
To compute the mixed strategy Nash equilibrium, assign A the probability p
of playing H and (1−p) of playing T, and assign B the probability q of
playing H and (1−q) of playing T.
E[payoff for A playing H] = (−1)q + (+1)(1−q) = 1−2q
E[payoff for A playing T] = (+1)q + (−1)(1−q) = 2q−1
E[payoff for A playing H] = E[payoff for A playing T] ⇒ 1−2q = 2q−1 ⇒ q = 1/2
We have calculated the strategy for player 2, but we haven't used their payoffs, only those for player 1! How is this possible?
Here comes a more or less complete citation of an earlier answer of mine:
This equilibrium can be found by noting that a strategy pair $(σ_1,σ_2)$ is a mixed strategy NE, iff every player is indifferent between the pure strategies played with positive probability in an equilibrium and each player weakly prefers the strategies played with positive probability to those played with zero probability.
Thus assuming player $j$ plays heads with probability $p$, player $i$ would obtain: $u(H,p)=p−(1−p)=2p−1$ playing $H$, and obtain $u(T,p)=−p+(1−p)=1−2p$ playing $T$. These two utilities are equal only if $p=1/2$. Assuming $i$ plays heads with probability $q$, by similar reasoning we obtain $q=1/2$.
Also as you noted above every finite game has a mixed strategy equilibrium.This can be shown by using correspondences (set valued functions) and kakutanis fixed point theorem. For an outline of the proof, you can have a look at the wikipedia page on Nash equilibrium. Although, to warn you, that site leaves out an tremendous amount of detail.