Maybe you can look at the proof of BC's lemma with the previous examples in mind.

"For any n" => intersection

"there exits" => union

and you can use the equality $\displaystyle P(A\cap B)=P(A)P(B)$ for independent events, and the inequality $\displaystyle P(A\cup B)\leq P(A)+P(B)$ in general.

Just an example. Translating into symbols, I suggested to first show that, for any $\displaystyle M\geq 1$, $\displaystyle P(\bigcup_{n\geq M}A_n)=1$. In order to use independence, I must handle intersections, hence let's write $\displaystyle P(\bigcup_{n\geq M}A_n)=1-P(\bigcap_{n\geq M}A_n^c)$. And we have $\displaystyle P(\bigcap_{n\geq M}A_n^c)=\prod_{n\geq M} P(A_n^c)=\prod_{n\geq M}\frac{1}{2}=0$ by independence.

(If you aren't used to applying independence to infinitely many events at once, you can say $\displaystyle P(\bigcap_{n\geq M}A_n^c)\leq P(\bigcap_{n=M}^{M+N-1} A_n^c)=\frac{1}{2^N}$ for all $\displaystyle N$, hence $\displaystyle P(\bigcap_{n\geq M}A_n^c)= 0$. )

Thus $\displaystyle P(\bigcup_{n\geq M}A_n)=1$. What about $\displaystyle P(\bigcap_{M\geq 0}\bigcup_{n\geq M}A_n)$? Switching to the complement: $\displaystyle P(\bigcup_{M\geq 0}\left(\bigcup_{n\geq M}A_n\right)^c)\leq $ $\displaystyle \sum_{M\geq 0}P(\left(\bigcup_{n\geq M}A_n\right)^c)=\sum_{M\geq 0} 0=0$ (using what we got before). In conclusion, $\displaystyle P(\bigcap_{M\geq 0}\bigcup_{n\geq M}A_n)=1$. This is what you need.

You can remember this (this was the last part): a countable intersection of almost sure events is almost sure.