We have to show that
$\qquad (E_1 + E_2).F^\omega = E_1.(F^\omega) + E_2.(F^\omega)$.
We have do to some preparation. $E_1, E_2$ and $F$ are regular expressions, so they have (deterministic) finite automata $A_1, A_2, A_F$. It is easy to see that we can construct NBA for the left and right hand side from those, respectively:

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Showing that these are correct should be a simple exercise, given the proof for the equivalence of $\omega$-regular expressions and NBA you should have seen.
Note that the transitions going into the DFA connect with their respective initial states, while the outgoing one originate from their final states (w.l.o.g. there is only one in each DFA).
Now, as usual, show $\subseteq$ and $\supseteq$ separately. Both directions work in the same way; let's look at it for $\subseteq$.
Assume a word $w$ is in the language generated by $(E_1 + E_2).F^\omega$. Then, there is an accepting (infinite) path in above automaton. The same path basically works for the lower automatong: depending on whether the run goes through $A_1$ or $A_2$, replace $q_m$ with $q_{m1}$/$q_{m2}$ and $q_e$ similarly.
A similar construction works for $\supseteq$.