Set of Infinite Sequences is Uncountable

Theorem
Let $S$ be a set which contains more than one element.

Let $S^\infty$ denote the set of all sequences of elements of $S$.

Then $S^\infty$ is uncountable.

Proof
As $S$ has more than one element, it must have at least two.

So, let $a, b \in S$ be those two elements.

Let $Z$ be the set of all sequences from $\left\{{a, b}\right\}$.

Suppose $S^\infty$ were countable.

From Subset of Countably Infinite Set is Countable, $Z$ is likewise countable.

So by definition, it would be possible to set up a bijection $\phi: Z \leftrightarrow \N$ between $Z$ and the set $\N$ of natural numbers:

where each of $z_{rs}$ is:
 * an element of $\left\{{a, b}\right\}$
 * the $s$'th element of $\left \langle {z_r} \right \rangle$, the particular sequence which is in correspondence with $r \in \N$.

Now we create the sequence $Q = \left \langle {q_{dd}}\right \rangle$ where:
 * $q_{dd} = \begin{cases}

a & : z_{dd} = b \\ b & : z_{dd} = a \end{cases}$

Thus:
 * $\forall n \in \N: q_{nn} \notin \left \langle {z_n} \right \rangle$

and so:
 * $\forall n \in \N: Q \ne \left \langle {z_n} \right \rangle$

So $Q \notin Z$.

That is, we have constructed a sequence which has not been placed into correspondence with an element of $\N$.

Thus by definition $Z$ is uncountable.

Hence, by the Rule of Transposition, $S^\infty$ is likewise uncountable.