How to use Borel-Cantelli specifically to show that the probability of a simple random walk returning to the origin in finite time is 1? Suppose we have that $X_i$ are iid random variables with $P(X_i =1) = P(X_i = -1) = 1/2$ and that $X_0 = 0$. Then, we define the simple symmetric random walk to be $S_n = \sum_{i=1}^n X_i$. We define the hitting time as $\tau_{i,j} = min\{n \geq 1 : S_n = j \ |\ S_0 = i\}$. I would like to show that $P(\tau_{0,0} < \infty) = 1$. 
Specifically, I would like to use the Borel-Cantelli lemma to show that if I can define some independent sequence of events here, and show that the sum of the events is infinite, then moves of some finite length happen infinitely often, and so I can show that it must return to $0$. However, I am not sure how to define such an indexed event. Could someone give me a hint? Thank you
 A: My reputation is under 50, hence I cannot add comments to contact the user jwg. So this is the only way to warn that, IMHO, the above solution of jwg is probably wrong. He states that

So the events that it reaches $n$ for the first time, and then reaches $0$
  before $n+1$, are disjoint and independent.

Yes, they are disjoint, which implies that $\sum_{n=1}^\infty P(A_n) = P(\cup_{n=1}^\infty A_n) \leq 1$, which contradicts

The sum of their probabilities is infinite...

Another problem is the independence:
Independence of, say, $A_2$ and $A_6$ would mean that $P(A_2 \cap A_6) = P(A_2) P(A_6)$. However, the left-hand side is zero (since $A_2$ and $A_6$ are disjoint), while the right-hand side is positive (since  $P(A_n)>0$ for all $n$).
It will be exremely enlightening, if jwg provides a more detailed reasoning. The possibility of proving the posted problem just by Borel-Cantelli makes me very curious, since I have never seen that kind of solution.
A: As pointed out in comments, there could be good ways of doing this using not Borel-Cantelli, but the properties of either Markov Chains or (I would recommend) martingales, since your random walk is both a Markov chain and a martingale.
However, there is a neat argument which uses the Borel-Cantelli lemma. Suppose that the random walk is at $n$. The probability of reaching $0$ before reaching $n+1$ is $\frac{1}{n+1}$.
For the random walk to go back to $0$, it must reach some maximum $k$, then go back $0$ before ever reaching $k+1$. (Or the same argument by symmetry if it goes to $-1$ before $1$.) So the events that it reaches $n$ for the first time, and then reaches $0$ before $n+1$, are disjoint and independent. The sum of their probabilities is infinite, hence at least one of them must occur with probability $1$.
To see that the probability, when at $n$, of reaching $0$ before $n+1$ is $\frac{1}{n+1}$, you could use a simple martingale argument. If win \$1 for heads and lose \$1 for tails, and stop gambling when you win \$1 or lose \$n, the probability of winning or losing must be such that your expectation is \$0.
One can also see that to remain in a range for ever without ever reaching either of the endpoints has probability $0$. To see this, just consider that there is a positive probability of getting $m$ heads or $m$ tails in a row, where $m$ is the length of the range. If we have infinite tries, we must, with probability $1$, get this sequence eventually.
