What you have to understand is holes don't actually exist, electrons are the only charge carriers in a semi-conductor but there are 2 types of electrons 1) free electrons and
2) bounded electrons.The "Electron flow" that is usually referred to in literature refers to the flow of free electrons and the "Hole flow" is the flow due to bounded electrons.
If a bounded electron moves in a particular direction it leaves a hole at the position it was previously at and it removes the hole at its new position.So it is equivalent to think of bounded electron flow as hole flow.
To understand the process better lets let b - bounded electron , h - hole and take the following illustration as a row of bounded electrons and a hole in a semi-conductor.
When there is a potential difference across semiconductor the bounded electrons will be attracted towards the positive terminal and so the bounded electron to the right of the hole will move left, this causes a hole to "move" right, the same process will repeat when the hole has moved to its new position, it will then seem as though the hole is the one moving.
$$
\text{+ b h b b b b -} \\
\text{+ b b h b b b -} \\
\text{+ b b b h b b -} \\
\text{+ b b b b h b -} \\
$$
Now with this knowledge in mind we can proceed to analyze a reverse biased semiconductor.Consider the following a negativily biased p-region (excess holes) next to a depletion region (all electrons bounded).
When a negative bias is applied on the p-type material holes will move left making the depletion region get bigger as it can be seen in this illustration.
originally:
$$
\text{ b h b b b h} \hspace{0.3cm} \text{b b b b} \\
\text{ b b h b h b} \hspace{0.3cm} \text{b b b b} \\
\text{ b b b h b h} \hspace{0.3cm} \text{b b b b} \\
\text{ b b b b h b} \hspace{0.3cm} \text{b b b b} \\
$$
then
$$
\text{- h b b b h b} \hspace{0.3cm} \text{b b b b} \\
\text{- b h b h b b} \hspace{0.3cm} \text{b b b b} \\
\text{- b b h b b b} \hspace{0.3cm} \text{b b b b} \\
\text{- b b b h b b} \hspace{0.3cm} \text{b b b b} \\
$$
then
$$
\text{- b b b h b b} \hspace{0.3cm} \text{b b b b} \\
\text{- h b b h b b} \hspace{0.3cm} \text{b b b b} \\
\text{- b h b b b b} \hspace{0.3cm} \text{b b b b } \\
\text{- b b h b b b} \hspace{0.3cm} \text{b b b b } \\
$$
note how the depletion region has now increased by 2 columns, this is a simplified illustration of what happens.Hope the diagrams were clear enough, I know they are very crude.
In a nutshell, bipolar junction transistors work because of the physical geometry of the two junctions. The base layer is very thin, and the charge carriers that are flowing from the emitter to the base do not recombine right away — most of them pass right through the base altogether and enter the depletion region of the reverse-biased base-collector junction. Once this happens, the strong field in this region quickly sweeps them the rest of the way to the collector terminal, becoming the collector current.
Best Answer
The resistance being "infinite" (its not actually, but its certainly very large) is a result of the DC behavior of the diode.
When you reverse bias a diode, the DC current flowing through it drops to basically zero. In the case of a PN diode, this is due to the lack of carriers flowing across the increased depletion region (until you reach breakdown). This is represented in the model by the resistor (a DC component) becoming infinite. However, as you said, the capacitance is still there and gets smaller because the depletion is wider (the AC component).
The AC impedance of the diode model never truly reaches infinity because of the capacitor, so to an AC signal there is still some current flowing in to the diode when it is reverse biased. If you think about it a bit, the current during reverse bias is produced by the depletion region expanding, a process that requires additional electrons or holes, but that current only occurs so long as the depletion region is actually changing. As it reaches its size for the applied voltage, the current will dwindle to near zero...similar to a capacitor charging. Hence, it can be represented by a capacitor in the model, whose capacitance is strong function of the applied voltage (which is very useful in oscillators since you can change the value of a component (the capacitor) simply by changing the voltage applied).
Normal ceramic capacitors also have a similar voltage-dependent effect, but I believe it's far weaker (requires larger voltage changes).