Proving that a set is countable 
My attempt is as follows:
We will start by considering each ‘walk’ as some set of steps. 
So a walk of length $m$ can be represented by the set $S_m =$ {$x_1, x_2, x_3, … , x_m$}. 
To each of these steps we assign at random a value of $0$ or $1$.
The number of possible walks of a certain length, say $n$, is the same as the number of functions from $S_n$ to {$0, 1$}, which is $2^n$. 
Now, the question tells us that the number of steps in a certain walk can get ‘very large’ so, assuming we begin the walk at the entrance to the pub, it would be silly to have walks consisting only of a ‘very large’ number of backward steps. It may, therefore, be better to write the number of possible walks of length $n$ as being $\leqslant 2^n$
The set of all possible random walks is then the disjoint union 
$S = \bigcup_{r = 1}^{L} S_r$, where $L$ is possibly a very large number, the cardinality of which is $\leqslant 2^1 + 2^2 + 2^3 + ... +2^L$ So the cardinality is bounded above by a finite quantity, so there is a bijection $f : \mathbb{N}_{i} \mapsto S$ for some $1 \leqslant i \leqslant 2^1 + 2^2 + 2^3 + ... +2^L$ hence the set is countable. 
The context is a little peculiar, so I'm not entirely convinced that my interpretation of the question is correct, so if anyone could give me some feedback on this that'd be great.
 A: It's a lot simpler than you seem to think. What we need to prove is that we can make a sequence of walks $x_n$ for $n \in \Bbb N$ such that any possible walk is somewhere in that sequence.
We start with the walk where he collapses immediately: $x_1 = \{\}$. Next come the two possible walks where he manages to stumble a single step before collapsing: $x_2 = \{0\}$, $x_3 = \{1\}$. Then comes the four walks consisting of two steps: 
$$
x_4 = \{00\},\quad x_5 = \{01\},\quad x_6 = \{10\},\quad x_7 = \{11\}
$$
and so on.
So why does this list contain every single walk? Simple, because for any given walk $x$, it has a finite number $n$ of steps. We include all possible walks of $n$ steps from $x_{2^n}$ to $x_{2^{n+1}-1}$, so in particular, $x$ has to be in there somewhere.
A: Your argument introduces $L$ but says nothing of how large $L$ might be.  In fact, the correct interpretation of the problem is that each random walk is finite, but there is no limit on how long the length may be.
So the question becomes "what is the cardnality of the set $S$ of all finite sequences $s_0s_1\ldots s_k$ of zero's and ones?"
Here is a $1:1$ mapping of $S$ to $\Bbb{Z^+}$.  The empty walk (length zero) maps to $1$ and for all other walks, 
$$s_0s_1\ldots s_k \mapsto 2^{k+1} + \sum_{n=0}^k s_n 2^{k-n}
$$ 
That is, just write a leading $1$ and treat the whole random walk sequensce as the rest of the bits in a binary number.  So the walk $1011$ maps to $11011 = 27$, for example.
Since there is a counting function, the set of all finite length random walks is countable.
A: Let a walk be marked as $X= ${$x_1, x_2........, x_m$} where $x_i$ are $0$ or $1$.
Let $f(X) = \sum x_i 2^i \in \mathbb N$.  
Suffices to prove $f$ is injective, that is, if $f(X) = f(W)$ then $X = W$.  This is precisely a matter of proving no natural number has binary expressions.  
Can you do that?
