You are making this problem a lot harder than it needs to be because the
random variables in question are two-valued, and the problem can be
treated as one of independence of events rather than independence of
random variables. In what follows, I will treat the independence of
events even though the events will be stated in terms of random variables.
Let $Z_0,Z_1,Z_2,\cdots$ be independent random variables $\ldots$
I will take this as the assertion that the countably infinite collection of events $A_i = \{Z_i = +1\}$ is a collection of independent events. Now, a countable collection of events is said to be a collection of
independent events if each finite subset (of cardinality $2$ or
more) is a collection of independent events. Recall that
$n\geq 2$ events $B_0, B_1, \cdots, B_{n-1}$ are said to be independent events
if
$$P(B_0\cap B_1\cap \cdots \cap B_{n-1})
= P(B_0)P(B_1) \cdots P(B_{n-1})$$
and every finite subset of two or more of these events is a
collection of independent events. Alternatively,
$B_0, B_1, \cdots, B_{n-1}$ are said to be independent events
if the following $2^n$ equations hold:
$$P(B_0^*\cap B_1^*\cap \cdots \cap B_{n-1}^*)
= P(B_0^*)P(B_1^*)\cdots P(B_{n-1}^*)\tag{1}$$
Note that in $(1)$, $B_i^*$ stands for $B_i$ or $B_i^c$
(same on both sides of $(1)$) and the $2^n$ choices
($B_i$ or $B_i^c$) give us the $2^n$ equations.
For our application, $A_i = \{Z_i = +1\}$ and $A_i^c = \{Z_i=-1\}$,
and so checking whether the $2^n$ equations
$$P(A_0^*\cap A_1^*\cap \cdots \cap A_{n-1}^*)
= P(A_0^*)P(A_1^*)\cdots P(A_{n-1}^*)\tag{2}$$
hold or not, is equivalent to checking that the
joint probability mass function (pmf) of $Z_0, Z_1, \cdots, Z_{n-1}$
factors into the product of the $n$ marginal pmfs at each and
every one of the points $(\pm 1, \pm 1, \cdots, \pm 1)$ which is
what you would be doing if you had never heard of independent
events, just about independent random variables.
Thus, the statement
Let $Z_0,Z_1,Z_2,\cdots$ be independent random variables $\ldots$
does mean, among other things, that $Z_0,Z_1,Z_2,\cdots, Z_{n-1}$
is a finite collection of independent random variables. But,
does the assertion
For all $n \geq 2$, $\{Z_0,Z_1,Z_2,\cdots, Z_{n-1}\}$ is a set
of $n$ independent random variables
imply that the
countably infinite set $\{Z_0,Z_1,Z_2,\cdots \}$ is a
collection of independent random variables?
The answer is Yes, because we know by hypothesis
that some specific finite
subsets of $\{Z_0,Z_1,Z_2,\cdots \}$ are independent random
variables, while any other finite subset, say $\{Z_2, Z_5, Z_{313}\}$,
is a subset of $\{Z_0, Z_1, \cdots, Z_{313}\}$ which are independent
per the hypothesis and so the subset is also a set of independent
random variables.
In your question, with each $a_i \in \{+1, -1\}$ and
defining $b_i = \prod_{j=0}^i a_j$ which is also in $\{+1,-1\}$,
\begin{align}
P(X_0 = a_0, X_1 = a_1, \cdots, X_n = a_n)
&= P(Z_0 = a_0, Z_1 = a_0a_1, Z_2 = a_0a_1a_2, \cdots, Z_n = a_0a_1...a_n)\\
&= P(Z_0=b_0, Z_1 = b_1, \cdots, Z_n = b_n)\\
&= \prod_{i=0}^n P(Z_i = b_i)\\
&= 2^{-(n+1)}\\
&= \prod_{i=0}^n P(X_i = a_i),
\end{align}
that is, all $2^{n+1}$ equations of the form $(2)$ hold.
Thus, for each $n \geq 1$, $X_0, X_1, \cdots, X_n$ are
independent random variables, and therefore the
countably infinite collection $\{X_0, X_1, \cdots\}$
of random variables is a collection of independent
random variables.
After reading over my revised answer, perhaps it is I who is
making the problem much harder than necessary. My apologies.
