Infinite Bernoulli Trials yielding the set of all infinite binary sequences of outcomes As I learned it, a Bernoulli trial has exactly two possible outcomes (sometimes symbolized by $1$ or $0$), with each trial being probabilistically independent and each trial (or "experiment") having the same probability; and a sample space is the set of all possible outcomes; e.g. flipping a fair coin twice would yield a sample space of $\{HH, HT, TH, TT\}$.
The question: how to mathematically prove that an infinite sequence of Bernoulli trials yields a sample space of the set of all infinite binary sequences of outcomes?
It seems obvious to me that an infinite sequence of Bernoulli trials yields a sample space that is the set of all infinite binary sequences of outcomes (or "Cantor space" as I was taught), but I've interacted with some who doubt it. For sake of concreteness, I'll use the classic example of flipping a fair coin for the Bernoulli trial, such that infinitely many of these Bernoulli trials involves flipping a fair coin infinitely many times, and the apparent sample space would be every possible infinite sequence of heads/tails.
I thought of using mathematical induction to show this, but the response I got from using this was that it applies only to any finite number but not the whole set. For example, if I wanted to show that a sequence of all heads was part of the sample space via mathematical induction, at most I could prove this (where $H(x)$ means every flip less than or equal to the $x$th flip came up heads):
$ \forall x \lozenge H(x)$
And not this:
$\lozenge \forall x  H(x)$
As such, I don't know to mathematically prove that an infinite sequence of Bernoulli trials yields a sample space of the set of all infinite binary sequences of outcomes. How does one do it?
To quote from this MIT open course link:

A Bernoulli process may be limited to a particular number of trials (e.g. 7 games, or 10 coin tosses), or it may go on indefinitely, in which case we may regard it as an infinite process. A finite Bernoulli sample space
consists of all binary sequences of some particular length $n$ (1 denotes
success, 0 failure). In the infinite case, the sample space consists of all infinite binary sequences.

The notion that, "In the infinite case, the sample space consists of all infinite binary sequences" is what I'm looking to prove.
 A: Let $(S, \mathcal{F}, P)$ be a probability triplet: $S$ is the sample space, $\mathcal{F}$ is a sigma algebra on $S$ (containing all the events), and $P:\mathcal{F}\rightarrow \mathbb{R}$ is a probability measure.
Suppose $\{X_i\}_{i=1}^{\infty}$ is a sequence of mutually independent and identically distributed (i.i.d.) random elements. In particular
$$X_i:S\rightarrow \{H, T\}$$
is a measurable map from the sample space to the set $\{H,T\}$, so for each $i \in \{1, 2, 3,...\}$ we have
$$\{\omega \in S: X_i(\omega) = H\} \in \mathcal{F}$$
Assume $P[X_i=H]=P[X_i=T]=1/2$.
Claim 1: For each sequence $\{h_i\}_{i=1}^{\infty}$ with $h_i \in \{H,T\}$, we have
$$\cap_{i=1}^{\infty} \{X_i=h_i\}= \cap_{i=1}^{\infty} \{\omega \in S: X_i(\omega)=h_i\} \in \mathcal{F}$$
Proof: Since $\mathcal{F}$ is a sigma algebra,  the countable intersection of events in $\mathcal{F}$ is in $\mathcal{F}$.  $\Box$
Claim 2: It is possible to construct $(S, \mathcal{F}, P)$, for which such i.i.d. random elements $\{X_i\}$ exist, in these example cases:
a) $S = \{red, blue\} \cup [0,1)$
b) $S = A$, where $A=\{(h_1, h_2, h_3, ...) : h_i \in \{H,T\}\quad \forall i \in \{1, 2, 3,...\}\}$.
c) $S =  A \setminus \{(T, T, T, T, ...)\}$.
In particular, examples (a) and (c) can be viewed as "counter-examples" to your claim that the sample space must contain all binary sequences of H/T.  The example (c) contains all binary sequences of H/T except for the all-Tails sequence $(T, T, T, T,...)$ (just start with the probability triplet in part (b) but throw away the probability-0 outcome of all tails).

Quick justifications for (a)-(c):
a) Use $P[\{red\}]=P[\{blue\}]=0$ and choose $\omega \in [0,1)$ according to the Borel sigma algebra and Borel measure.  Write $\omega\in [0,1)$ as
$$ \omega = \sum_{i=1}^{\infty} \omega_i 2^{-i}$$
where $\{\omega_i\}$ is the unique binary expansion that does not contain an infinite tail of 1s.  Define for each $i \in \{1,2,3,...\}$
$$X_i(red)=X_i(blue)=H$$
For $\omega \in [0,1)$ define
$$X_i(\omega) = \left\{\begin{array}{cc}
H & \mbox{if $\omega_i=1$}\\
T & \mbox{else} 
\end{array}\right.$$
b) This is the standard one.
c) Just start with (b) and throw away an outcome of probability 0.
A: If we define a sample space as the set of all possible outcomes, an infinite sequence of Bernoulli trials implies a sample space of the set of all infinite binary sequences of outcomes given a few assumptions.
For the Bernoulli process in question: each trial has exactly two possible outcomes, each trial is mutually independent (in the sense that the probability of the outcome is unaffected by the outcomes of other trials), and the probability is the same in each trial. Some additional assumptions to make the reasoning a bit easier to follow if nothing else: each candidate outcome for a trial (for example, heads or tails) has a nonzero probability, and any outcome with a nonzero probability is possible.
Let $\Omega$ represent the sample space of an infinite sequence of Bernoulli trials, which is the set of all possible outcomes for that sequence. For concreteness, let's symbolize the sequence of Bernoulli outcomes as $H$ and $T$ ("heads" and "tails," respectively) like so:
$(a_{i})^\infty_{i = 1}  \text{where } a_{i} \in \{H, T\}$
Since $H$ and $T$ both have nonzero probability for each trial, and nonzero probabilities entail possibility:
$\forall i \in \mathbb{N} \left [ \lozenge (a_{i} = H) \land \lozenge (a_{i} = T) \right ]$
Given the aforementioned conditions (e.g., both $H$ and $T$ have a nonzero probability), the fact that the probability of a given trial’s outcome is unaffected by the outcomes of other trials entails that which outcomes are possible for each trial is also unaffected by the outcomes of other trials. For each trial: (a) both outcomes are possible; and (b) which outcome is possible is unaffected by the outcomes of other trials. Since every element in the sequence has the property of heads or tails both being possible values regardless of the values of the other sequence elements, this permits any infinite sequence of binary values.
To illustrate, can the first coinflip be heads? Yes, since for each sequence element, $H$ or $T$ is a possible outcome, which would include the first coinflip. Given that the first coinflip is heads, can the second coinflip also be heads? Yes, since the possible outcomes of the second coinflip is unaffected by the outcomes of other coinflips; hence both $H$ and $T$ can be used here. If the first two coinflips are heads, can the third one be also? Yes, because the possible outcomes of the third coinflip is unaffected by the outcomes of the other coinflips, hence both $H$ and $T$ can be used here, and so on ad infinitum for all of the remaining sequence elements. And since the outcomes of heads is arbitrary here (e.g., we could just as well have used tails, heads, tails for the first three sequence elements, and then choose whatever binary sequence we wish for the remaining sequence elements) this can be generalized so that the sample space consists of all infinite binary sequences.
$\Omega = \left\lbrace  (a_{i})^\infty_{i = 1} \vert a_{i} \in \{H, T\} \right\rbrace$
