Proof that $P(A_{n}\,\text{i.o.})=1$ implies $P(\cup_{i=1}^{\infty}A_{i})=1$

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There are many, many proofs on MSE that show the "hard" direction in the proof that when $A_{1},A_{2},\cdots$ are mutually independent sets with $P(A_{i})<1$ that $P(A_{n}\,\text{occurs i.o.})=1$ if and only if $P(\cup_{i=1}^{\infty}A_{i})=1$.

I.e., there are a lot of proofs showing that $P(\cup_{i=1}^{\infty}A_{i})=1 \, \implies \, P(A_{n}\,\text{occurs i.o.})=1$. But, I am having trouble with the other direction, which according to a lot of people on here is "obvious". However, it's not obvious to me.

I am trying to show that $P(A_{n}\,\text{occurs i.o.})=1 \implies P(\cup_{i=1}^{\infty}A_{i})=1$, but I am getting stuck and am not sure how to finish my proof:

Suppose $P(A_{n}\,\text{occurs infinitely often})=1$. This is equivalent to saying that $P(\limsup_{n \to \infty}A_{n})=1 \Longleftrightarrow P(\cap_{n=1}^{\infty}\cup_{i=n}^{\infty}A_{i})=1 \Longleftrightarrow P(\lim_{n \to \infty}\cup_{i=n}^{\infty}A_{i})=1 \Longleftrightarrow \lim_{n \to \infty}P(\cup_{i=n}^{\infty}A_{i}) = 1$

But, I'm not sure where to do from here to get $P(\cup_{i=1}^{\infty}A_{i})=1$.

Could somebody help me fill in the gaps I need in order to finish it?

Thank you for your time and patience!

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By definition of intersection: $$\bigcap_{n=1}^{\infty}B_n\subseteq B_1$$

Applying this we find:

$$\bigcap_{n=1}^{\infty}\bigcup_{i=n}^{\infty}A_i\subseteq\bigcup_{i=1}^{\infty}A_i$$

and consequently $$P\left(\bigcap_{n=1}^{\infty}\bigcup_{i=n}^{\infty}A_i\right)\leq P\left(\bigcup_{i=1}^{\infty}A_i\right)\leq 1$$

So if LHS takes value $1$ then also RHS takes value $1$.