This solution is entirely due to zeb, and is paraphrased from this xkcd thread post.
Let $\epsilon>0$ be a small number. Take a particular maze, and imagine a robot who starts on the starting northwest square, then moves in a random direction once per minute according to the below probabilities:
^ e
|
e <--+--> 1/2 - e
|
v 1/2 - e
Notice that the robot will tend to the southeast over time, but still work its way around obstacles.
The robot's position will be a Markov chain, whose states are the squares reachable from the start square. Using Markov chain theory, you can show the probability distribution of the robot's location will approach a unique stationary distribution as the number of steps increases. Let $p(i,j)$ be this limit distribution, where $(1,1)$ is the starting square, and $(8,8)$ is the finish.
An amazing fact is this: no matter how the maze is structured, the limit distribution $p(i,j)$ will be of the form $$p(i,j)=K\left(\frac{\frac12-\epsilon}{\epsilon}\right)^{i+j},$$where $K$ is a normalizing constant. This is quite surprising: no matter how snakey the maze is, over time, the probability this robot will be on a square only depends on which diagonal that square is on. To check this, you need only show that this distribution is a fixed point of the Markov process. Namely, show that if the robot has this distribution, then one minute later, it will still have this distribution. I omit the proof.
In this distribution, $p(8,8)$ has the largest probability, and every other probability is at most $\frac{\epsilon}{\frac12-\epsilon}\cdot p(8,8)\le 4\epsilon$, implying that $p(8,8)\ge 1-63\cdot 4\epsilon$. Thus, if we choose $\epsilon$ small enough, we can make $p(8,8)$ as close to $1$ as we want.
Finally, imagine if we set up every possible maze, choose a random program in this fashion of length $L$, then run the same program on a robot placed on the start square of each maze. By the previous discussion, we can choose $\epsilon$ small enough, and $L$ large enough, so that each robot has a less than $\frac1{2^{112}}$ probability of not ending on $(8,8)$. Since there are less than $2^{112}$ mazes, we have that
$$
\begin{align}
P(\text{all robots end on finish})
&=1-P(\text{at least one robot doesn't})\\
&\ge 1-\sum_{m=1}^{\text{# of mazes}} P(\text{robot on maze m doesn't end on finish})\\
&>1-2^{112}\cdot\frac1{2^{112}}=0 \\
\end{align}
$$
Thus, there is a nonzero probability of a random program succeeding on all mazes, so there must exist a particular, determistic program which succeeds for all mazes.
