I'm supposed to show this with martingale convergence thm. I have tried setting up one barrier at a fixed $n$. And the martingale convergence says it will converges to a random variable almost surely. But why would that imply it will hit the barrier?
EDIT: I think I just came up with a way to show this as I'm typing (setting up another barrier and gradually moving it to infinity). But I'd still like to know possible, more straightforward alternatives.
Take $M_n$ to be the location after $n$ steps, and $T$ to be the first time at which our random walk is equal to 1. Then, $X_n = M_{\min(T,n)}$ forms a Martingale, and since the $X_n$ are bounded above by 1 by Martingale convergence will converge a.s. to some random variable $X$.
We want to show that $X=1$. Off the top of my head I can't think of any elegant way to do this, but we can observe that since $X,X_n$ are integer valued the PDFs of $X_n$ converge pointwise to the PDF of $X$. One can then check that this forces $X=1$. (since $P(X_{n+1}=1) = P(X_n=1) + \frac{1}{2}P(X_{n}=0)$ and $|P(X_{n+1}=1)- P(X_n=1)| \to 0$, this forces $P(X_n=0)\to 0$, and one can argue onwards to show that $P(X_n= k)=0$ for any negative $k$.)
This tells us that our original random walk $M_n$ hits 1 almost surely, which implies that we hit any integer point almost surely.