There are infinite sequences of coin flips that do not contain a single stretch of 100 consecutive heads. In fact, there are uncountably infinitely many: let $t_1,t_2,t_3,\dots$ be an infinite sequence of numbers from $\mathbb{N}$, and $h_1,h_2,h_3,\dots$ be an infinite sequence of numbers from $\{0,1,2,\dots,99\}$. Then a sequence of $t_1$ tails, $h_1$ heads, $t_2$ tails, $h_2$ heads, and so on is a sequence with no stretch of 100 heads.
When dealing with infinity, "almost surely" deals with situations that occur with probability $1$ according to the formal notions associated with a probability space. This does not imply there are no valid situations where an event occurs or a hypothesis is sasisfied, as Wikipedia says:
If an event is sure, then it will always happen, and no outcome not in this event can possibly occur. If an event is almost sure, then outcomes not in this event are theoretically possible; however, the probability of such an outcome occurring is smaller than any fixed positive probability, and therefore must be 0. Thus, one cannot definitively say that these outcomes will never occur, but can for most purposes assume this to be true.
Intuitively, $A \in \chi = \bigcap_{n} \mathbb{F}^{\infty}_{n}$ means $A$ is an event in the limiting random variable, i.e. $\lim_{n\rightarrow\infty}X_n$. Hence it is not affected by finite permutations, and is an exchangeable/symmetric event.
A precise proof is mostly about definition understanding and manipulation. Let $\mathcal{R}$ be Borel sets of real number $R$. $(R, \mathcal{R})$ is a measurable space.
An event $A \in \mathbb{F}^{\infty}_{n}$, by definition, is equivalent to $A\equiv\{\omega \in \Omega: X_i(w) \in B_i, B_i \in \mathcal{R} \text{ for }i \ge n\}$. Note the definition of $A$ has no requirements on $X_i(\omega)$ with $1 \le i < n$, and they can be any real values. We can equivalent write $B_i=R$ for $1 \le i < n$ in this case.
Furthermore, $\pi A \equiv \{\omega \in \Omega: X_{\pi(i)}(w) \in B_i \text{ for }i \ge 1\}. $ $B_i$s here are from the definition of the event $A$.
For any finite permutation $\pi$, there exists a $N$ so that for all $n \ge N$, $\pi_(n)=n$. Since $A \in \chi$, $A \in \mathbb{F}^{\infty}_{N}$. We have
$$\pi A = \{\omega \in \Omega: X_{\pi(i)}(w) \in B_i \text{ for }i \ge 1\} \\ = \{\omega \in \Omega: X_{\pi(i)}(w) \in B_i\text{ for }1 \le i < N, X_{\pi(i)}(\omega) \in B_i \text{ for }i \ge N\} \\= \{\omega \in \Omega: X_{\pi(i)}(w) \in R\text{ for }1 \le i < N, X_i(\omega) \in B_i \text{ for }i \ge N\} \\= \{\omega \in \Omega: X_i(\omega) \in B_i \text{ for }i \ge N\} = A.$$
Best Answer
This is simple application of Borel Cantelli Lemma. If $\sum P(X_n \leq a_n)=\infty$ then $X_n \leq a_n$ holds infinitely often with probability $1$. So you only have to check that $\sum [1-e^{-(\ln n+ \ln \ln n +\ln^{2} \ln \ln n)}]=\infty$. Can you check this? [The general term of this series does not tend to $0$].