Let $U = $ the set of all infinite series
$S = $ the set of all infinite series with infinite 1s and 0s
$A = $ the set of all infinite series with finite 0s
$B = $ the set of all infinite series with finite 1s.
$U = S \cup A \cup B$.
Claim 1: U is uncountable.
Claim 2: A and B are countable
Claim 3: the countable union of countable set is countable.
Hence, if $S$ were countable, $U$ would be countable. It isn't so $S$ is uncountable.
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Claim 1: This is Cantor's diagonal. If U were countable we could list all sequences so $U = \{A_i\} = \{\{a_{i_k}\}\}$. Then the sequence $B = \{1-a_{i_i}\} \not \in U$. A contradiction so $U$ is uncountable.
Claim 2: If an infinite sequence, {$a_n$} contains a finite number of 0s then there is an $a_m$ that is the last 0 value and all the remaining values are 1. Consider the finite sequence {$b_n$} that has $m$ terms were $b_i = a_i$. For every infinite sequence with finite number of 0 there is such a corresponding finite sequence. And for every finite sequence we can find a corresponding infinite sequence by "sticking" an infinite number of 1s on the end.
So there is a 1-1 correspondence between finite sequences and infinite sequences with finite 0s.
But the set of finite sequences is countable. ($t(\{b_n\})=\sum_{i=0}^m b_n*2^i \in \mathbb N$ is a 1-1 mapping from the finite sequences to the natural numbers.)
So there are countably many infinite sequences with finite 0s, and likewise there are countably many infinite sequences with finite 1.
Claim 3: Let $\{S_i\}$ be a class of countable sets so that each $S_j = \{s_{j,k}\}$ so $\cup_i S_i = \{s_{j,k}\}$ with some possibility of multiple representation. It suffices to show that $\mathbb N \times \mathbb N$ is countable. Basically the map (0,0)-> 0, (1,0) ->1, (0,1) ->2, (2,0)->3, (1,1) -> 4, (0,2)->5 etc or $(i,j) \rightarrow \sum_{l=0}^{j+k}l + k$ is a 1-1 mapping.
Best Answer
You were supposed to be assuming, for the sake of contradiction, that $S$ was countably infinite; Cantor's diagonal argument tells you how to construct, for any countably infinite collection of binary sequences, a binary sequence not in that collection. Thus, the set $S$ of all binary sequences (which is a perfectly well-defined object) is uncountable.