I think WalrusMcFishSr's post is better, but here's an argument I thought up. (PPE: I think it's actually the same as the argument in the link, now that I've read the pdf more closely)
Think of it as a random walk, where we either travel +1 or -1 with equal probability. The question becomes P(visit +1 at any point, starting from state 0)
Let that be p. If the first trial is +1, then we've obviously made it, so
P(visit +1 at any point, starting from state 0) = 1/2 + 1/2 * P(visit +1 at any time, starting from -1)
To visit +1 starting from -1, we must first pass through 0, so
P(visit +1 at any time starting at 0) = 1/2 + 1/2 * P(visit 0 at any time starting from -1) * P(visit +1 at any time starting from 0)
The probabilities are the same under translation (we only care about walking right by 1, not where we start from), so this turns into
p = 1/2 + 1/2 * p^2
which has a solution at only p = 1
Edit: yep it for sure is the same argument. I also only solved it for when the random walk chance is 1/2, so you should just look at the pdf instead because that solves it generally.