This answer reminds me of how I was taught limits the first time around, which was a kind of middle ground between "total handwaving" and rigorous $\epsilon,\delta$ definitions. Namely I was told, only in words, something along the lines of "$f(x) \to L$ as $x \to a$ if you can make $f(x)$ arbitrarily close to $L$ by taking $x$ arbitrarily close to $a$". This is not 100% unambiguous, but I found it both enlightening and intelligible at the time.

It can help with your intuition all around. The whole point of these conditions is that, loosely speaking, the boundary does not have any "corners" so "sharp" that a Brownian motion can never find its way to them without hitting some other point on the boundary first. Classical potential theory has a different way of looking at this same notion of "sharpness" which may help you with intuition (though it will probably not help you with merely carrying out this exact proof, which as far as I can tell is merely a matter of technicality).

It may help with your intuition to look into classical potential theory, which has its own (non-probabilistic) mechanisms for solving the exact same problem.

@SergueiPopov That at least helps with perspective, because for that $V$ you have $\pi(x)^{-1}=E[\tau_{xx}] \leq E[\tau_{x x_0}]+E[\tau_{x_0 x}]=V(x)+E[\tau_{x_0 x}]$ so that $V(x) \geq \pi(x)^{-1}-E[\tau_{x_0 x}]$. But I suspect this bound is frequently trivial in practice. Do you know any techniques in this general area (constructing Lyapunov functions for complex chains)? (It may be that I just asked the wrong question.)

To save us some time, maybe move point #5 closer to the top of the question, since that was my first reflex.

@AryehKontorovich In an infinite dimensional Banach space, an open ball is not precompact, and the support of a nonzero continuous function contains some open ball, so it is not compact. I'm not sure what separable had to do with anything.

What kind of additional hypotheses might you hope for? For example I think it is not hard to get such a result if $g$ is uniformly bounded below (away from zero) and $|f|$ is uniformly bounded above.

You'd have the same problem if you included weak derivatives up to order $k$ in there, you'd get $W^{k,2}$ instead of $W^{k,p}$ which would be weaker since the domain is bounded.

One way would be to artificially stop the process at a boundary to the left of $x(0)$, solve that PDE, and then send that boundary to $-\infty$. You can get some intuition for this trick from gambler's ruin. Consider studying the time to hit zero in a simple (possibly asymmetric) random walk on $\mathbb{N}_0$. You can understand this quantity by artificially deciding that the gambler will quit if he earns some amount $n$ (which gives two boundary conditions for a second order recurrence, as you should have) and then make the gambler "infinitely greedy" by sending $n \to \infty$.