Walking to infinity stepping on randomly selected lattice points Suppose you randomly fill the infinite non-negative quadrant of $\mathbb{Z}^2$ with $1$'s and $0$'s, with $1$ occurring with probability $p$ (and $0$ with probability $1-p$).
The lowerleft corner of the quadrant is at the origin $(0,0)$, marked red below:
$$
\left[
\begin{array}{ccccc}
 1 & 1 & 1 & 1 & 1 \\
 1 & 1 & 0 & 0 & 0 \\
 1 & 1 & 1 & 1 & 1 \\
 0 & 1 & 1 & 0 & 1 \\
 \color{red}{1} & 1 & 1 & 0 & 1 \\
\end{array}
\right]
$$

What is the probability that you can walk to $\infty$ starting
  at the origin, stepping only on $1$'s, moving one step N, S, E, or W at a time?

It cannot be $1$ regardless of the value of $p$, because there is always a $(1-p)$ chance that the origin itself is assigned $0$ and your walk cannot begin.
In the example above (which I generated with $p=0.7$), you can walk to the
boundary of the $5 \times 5$ snippet shown. But already with $50 \times 50$
and $p=0.7$, rarely can the boundary be reached.
Is the answer zero independent of how close $p$ is to $1$?
That would be somewhat counterintuitive...
 A: If there's no path to infinity, then there is a contiguous wall of zeros connecting the $x$-axis and the $y$-axis, with diagonal adjacency allowed. The probability of such a wall existing is bounded from above by the sum of the existence probabilities of all possible walls. Any wall starts at some point $n\ge0$ on the $x$-axis and traverses at least $n+1$ zeros before reaching the $y$-axis. In each of the at least $n$ steps, it has $5$ directional options, since if it retraces itself or moves diagonally backwards we could streamline it, and the initial step also has at most $5$ directions to go. So the sum of all existence probabilities is at most
$$
\sum_{n=0}^\infty\sum_{k=n}^\infty5^kq^{k+1}=\frac1{1-5q}\sum_{n=0}^\infty5^nq^{n+1}=\frac{q}{(1-5q)^2}\;,
$$
where $q=1-p$. Equating this to $1$ yields $q=(11-\sqrt{21})\,/\,50\approx0.12835$. But we can do better than that: For any given $m$, there is a finite probability for the path to traverse all walls up to $m$ steps, and thus a finite probability to reach infinity unless there is a wall with at least $m$ steps. Thus we can consider
$$
\sum_{n=m}^\infty\sum_{k=n}^\infty5^kq^{k+1}=\frac1{1-5q}\sum_{n=m}^\infty5^nq^{n+1}=\frac{q(5q)^m}{(1-5q)^2}\;.
$$
This is less than $1$ for sufficiently large $m$ if $q\lt0.2$. Thus $p^*\le0.8$, as you guessed.
We can do yet better by taking into account that after an axis-parallel step there are only $3$ different directions for the wall to take without allowing us to streamline it. Denoting by $a_j$ and $b_j$ the number of walls with $j$ steps that end and don't end in an axis-parallel step, respectively, we have
$$
\pmatrix{a\\b}_{j+1}=\pmatrix{1&2\\2&3}\pmatrix{a\\b}_j\;.
$$
The greater eigenvalue of this matrix is $2+\sqrt5\approx4.236$, so we should have
$$p^*\le1-\frac1{2+\sqrt5}=3-\sqrt5\approx0.764\;.$$
Here's code that estimates the expected Manhattan distance $d$ reachable from the origin. Here's a log-log plot of the results for $0.5\le p\le0.591$:

The $x$-axis shows $-\log(\tilde p_{\mathbb Z^2}-p)$, where $\tilde p_{\mathbb Z^2}=0.5927460507921$ is the best known estimate of the site percolation threshold for the square lattice, and the $y$-axis shows $\log\langle d\rangle$. The roughly linear relationship suggests $p^*=p^*_{\mathbb Z^2}$.
Here's code that estimates the probability to reach $x=50$ or $y=50$; here's the result:

The red data points show the results of the simulation; the green curve is the crude approximation
$$
p\cdot\frac{p^2}{p^2+(1-p)^2}\;,
$$
where the factor $p$ is for the origin itself to be a $1$ and the other factor is the probability that a single $1$ leads to two $1$s before it leads to two $0$s.
The results are clearly compatible with $p^*=p^*_{\mathbb Z^2}$. At $p=0.7$, the boundary is reached roughly half the time. As you say it can rarely be reached, there might be a bug in your code.
