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

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.
