PN junction: not sure to understand why reverse-biais lead to negligible current

I think you understand almost everything, but it seems the crucial thing you’re missing is this:

There will be a short-lived initial current but it will quickly (mostly) stop as soon as the potential at the top end of each wire (mostly) equilibrates with the potential at the bottom end of each wire. This never happens in the case of a resistor.

Let me explain how this happens, and why the resistor case is different. First, to make sure my makeshift terminology is clear, let's look at the picture:

enter image description here

When I say the top end of the left wire, I am referring to the place where the left wire connects to the p-type semiconductor. The bottom end is the place where it connects to the battery.

Now, to avoid a 0/0 indeterminacy, and to be more realistic, we should remember that each wire has some low, but non-zero resistance. This will actually make the analysis more clear. Also, let's only look at the left wire. The analysis for the right wire will be the same.

When you connect a battery to the diode, or to a resistor, what happens? We can think of the battery as some black box mechanism to enforce a constant potential difference $\Delta V$ between its negative and positive terminals, no matter what happens. This will set up a gradient of potential throughout the whole circuit, which will push the current to start flowing counterclockwise.

Let's focus again specifically on the left wire. As part of that overall gradient of potential, the battery created a potential difference between the two ends of the left wire, and the current is flowing from its top to its bottom end.

In the case of a resistor, as we know this will be the case forever, the current will never stop.

By contrast, in the case of the diode, as holes flow through the left wire, the left side of the junction is getting robbed of more and more holes, and will therefore develop a negative charge, which will gradually make it harder and harder for the holes to keep flowing. The same of course will happen in the wire on the right hand side, for electrons.

From the perspective of the potential, the potential at the top end of the left wire will get lower and lower until it (mostly) matches the potential at its bottom end, at which time current will (mostly) stop.

What's the crucial difference between the diode and the resistor? The holes have a very hard time crossing from the right side of the diode to the left side, to replenish the deficit! Why? If we look at the picture again, it would seem that the electric field in the depletion zone should help them cross from right to left, because it’s pointing to the left. So why wouldn't they want to cross?!

The answer is, if any hole makes it to the right side of the depletion zone, then yes, the electric field will propel it further to the left and it will happily cross. The problem is that the overwhelming majority of holes supplied by the right wire "die" way before they reach the depletion zone. And when I say die, I mean they will encounter an electron and annihilate.

The reason I say "(mostly)" several times in the above paragraphs is because there will be some very small number of holes that manage to make it from the right side of the diode to the left side, so some very small amount of current will exist. Even though the right side of the diode is teeming with electrons (majority carriers), thermal fluctuations occasionally produce an electron-hole pair (an electron gets excited from the valence band to the conduction band). If a hole produced this way makes it to the depletion zone without being annihilated (or is born in the depletion zone itself), then it will happily be swept away by the electric field and will make it to the left side of the diode to join its comrades. The tiny current produced this way is called the thermal current.

Note: just to make sure we understand the final configuration, and ignoring the tiny thermal current, in the end the potential everywhere in the left wire will be the same, but it will be lower than the potential everywhere in the right wire, by $\Delta V$, the battery’s voltage.