Showing that $P(E \cup F) + P(E \cup F^c) + P(E^c \cup F) + P(E^c \cup F^c) =3$. I need to prove the following:

For events $E$ and $F$,
$$P(E \cup F) + P(E \cup F^c) + P(E^c \cup F) + P(E^c \cup F^c)=3.$$

Is this proof sufficient?
By inclusion-exclusion principle:
$$P(E \cup F) + P(E \cup F^c) + P(E^c \cup F) + P(E^c \cup F^c)=$$
$$P(E)+P(F)-P(EF) + $$
$$P(E)+P(F^c)-P(EF^c) + $$
$$P(E^c)+P(F)-P(E^cF) + $$
$$P(E^c)+P(F^c)-P(E^cF^c) + $$
Since $P(A^c) = 1- P(A)$,
the previous term equals:
$$P(E) + P(F) - P(EF)+$$
$$P(E) + 1-P(F) - P(EF^c)+$$
$$1-P(E) + P(F) - P(E^cF)+$$
$$1-P(E) + 1-P(F) - P(E^cF^c)$$
$$=$$
$$4-P(EF)-P(EF^c)-P(E^cF)-P(E^cF^c) = 4-(P(EF)+P(EF^c)+P(E^cF)+P(E^cF^c))$$
Proof that $P(EF)+P(EF^c)+P(E^cF)+P(E^cF^c) = 1$:
$$P(EF) + P(EF^c) = P(E)$$
$$P(E^cF) + P(E^cF^c) = P(E^c)$$
Note that $EF$ and $EF^c$ are mutually exclusive.
And $E^cF$ and $E^cF^c$ are mutually exclusive.
$$P(E) + P(E^c) = P(E \cup E^c) = 1$$
Everything together:
$$4-1 = 3$$

Given a probability space $\Omega$, every $\omega \in \Omega$ is a member of exactly 3 among the 4 sets
$$ \tag{*} E \cup F, \quad  E \cup F^c, \quad  E^c \cup F, \quad  E^c \cup F^c \,.$$
E.g. if $\omega \in E \cap F$, then $\omega$ is contained in the first 3 sets in $(*)$, but not in the fourth. The other cases are similar. Thus, in a discrete probability space, adding the probabilities of the events in $(*)$ will yield
$$\sum_{\omega \in \Omega} 3 P(\omega) =3 \,.
$$
In  a general probability space, the sum of  the indicators of the events in $(*)$ is identically 3; now take expectations.
