I'm trying to prove that the infinite set of all words, meaningful or not, of any finite length, made with the 26 letters of the English alphabet is uncountable.
I believe the answer to be no but I am stuck trying to prove that one of the subsets is uncountable and thus the full set is too. Any hints would be greatly appreciated.
In fact, the set is countable. Let's replace the alphabet with the set of numbers $B=\{0,1, 2, 3, ..., 25\}$ - this has the same number of elements, so the problem is the same.
Now, given a string $\sigma=b_1b_2...b_n$ from $B$, consider the number $$1b_1b_2...b_n$$ where the $b_i$s are interpreted as base-$26$ digits; e.g. the string "$22$" gets sent to the base-$26$ number "$122$", or in base $10$ the number $$1\cdot 26^2+2\cdot 26+2=676+52+26=754.$$
This process of taking a string and outputting a number gives an injection from the set of finite strings of symbols in $B$, to the natural numbers; hence the set of finite strings of symbols in $B$ is countable.
We put "$1$" at the front to make sure that the map is injective - e.g. otherwise the strings "$034$" and "$34$" both yield the same number, whereas $1034$ and $134$ are very different numbers.
More generally, by the same trick the set of finite strings of symbols in $X$ for any finite $X$ is countable. And in fact more is true - as long as $X$ is countable, the set of finite strings of symbols in $X$ is countable, although this takes a bit more work to show (show that $(i)$ any finite power of a countable set is countable and $(ii)$ the union of countably many countable sets is countable (this is really implicit in $(i)$ - take the power to be $2$). Now the set of finite strings of symbols in $X$ can be decomposed into the sets of strings of length $n$, as $n$ varies through the naturals. Each of these sets is countable by $(i)$, and so by $(ii)$ the union of them all is countable.
By contrast, everything goes to heck if we look at infinite sequences instead. If $X$ has more than one element, then the set of infinite strings of symbols in $X$ is always uncountable. This is due to Cantor's diagonal argument, and shows that we have to be very careful when thinking about infinity to distinguish between finite and infinite strings, sets, and so on.
This set will be countable. Let $A$ be the set of 26 letters and $A^n$ be the set of words with $n$ letters, then $A^n$ is finite. Then your infinite set of all words is the union of the sets of words with length $n$: $$\bigcup_{n \in \mathbb{N}} A^n$$ hence a countable union of finite sets, hence countable.