Question: Let ${A\_n}$, n = 1, 2, 3, ... be a sequence of countable sets, and put $B = A\_1 \times A\_2 \times \cdots \times A\_n \times \cdots$. Show that B is uncountable. Prove that the same statement holds if each $A\_n$ = {0, 1}. 

I'm not exactly sure how I should begin this. Any help would be appreciated.

Answer: 
Assuming that for you *countable* means *in bijection with the set of natural numbers*, as each $A\_{n}$ is countable, $\vert A\_{n}\vert=\vert\mathbb{N}\vert$. Therefore one has

$$\vert B\vert = \vert\prod\_{n\in\mathbb{N}}A\_{n}\vert = \vert\prod\_{n\in\mathbb{N}}\mathbb{N}\vert =\vert\mathbb{N}^\mathbb{N}\vert\geq\vert 2^{\mathbb{N}}\vert =\vert\mathbb{R}\vert > \vert\mathbb{N}\vert ,$$

where the last equality is due (for example) to the binary representation of real numbers. This chain of inequalities proves both your statements.
Suppose $B$ is countable, then let $b\_n$ enumerate $B$. Define $d$ as follows:

If $b\_n(n) = a\_n\in A\_n$ then pick any $a'\_n\in A$ not equal to $a\_n$. Then $d(n) = a\_n'$.

$d$ must be an element of $B$ as it's $n$th componant is an element of $A\_n$, however $d\neq b\_n$ for each $n$, as $d$ disagrees with $d\_n$ on the $n$th place. Hence $d$ is not in the enumeration of $B$. A contradiction.

Hence $B$ is uncoutable.