I am reading this very interesting question and the answers below it... and two things amaze me:
First, how no one (I do not mean OP) evaluated these comprehensive answers with at least one elementary +1 (especially, the Tony's answer accompanied by a simulation). SE EE only loses from such an attitude towards else's achievements because, in this way, it will gradually turn from a cohesive society into a mechanical mixture of unrelated individuals.
Second, how did no one (excluding, to some extent, the DKNguyen's answer) give a simple and clear intuitive explanation of this circuit trick based only on basic electrical concepts?
The question is about the well-known circuit concept of "voltage shifting"... or, more precisely, "dynamic voltage shifting". We can see it, for example, in the bias circuits of AC amplifiers (decoupling capacitors), RC differentiating circuits, capacitive voltage multipliers, etc.
The best way to understand and explain it is to think of the capacitor (with a high enough capacitance) as a "rechargeable battery". In the OP's circuit, it is initially charged by the voltage source through the two (1k and 10 k) resistors in series to the magnitude of Vcc. But why do we need two resistors instead of only one?
The role of resistors is to decouple the capacitor from both source terminals. Thus, the fully charged capacitor behaves as a floating "battery" that can be easily "moved" in any direction by fixing one of its terminals to some of the source terminals. The resistors will not affect the voltages but only the common current. Here are the possible connections:
If we connect the positive capacitor terminal to the positive source terminal (turning on a switch connected between them), or the negative capacitor terminal to the negative source terminal, nothing (neither current or voltage) will change. The reason of that is because two equal voltage sources are connected in series and they neutralize each other. Their voltages are subtracted and the resulting voltage in the loop is zero. No current flows through the resistors... and if we short one of them, nothing will change.
Then, if we connect, according to the OP's question, the positive capacitor terminal to the negative source terminal (turning on the switch in the OP's figure), the negative capacitor terminal will be "shifted down" with Vcc.
Finally, if we connect the negative capacitor terminal to the positive source terminal, the positive capacitor terminal will be "shifted up" with Vcc… and its voltage (in respect to ground) will be 2Vcc. This means that the two voltage sources are connected in series in the same direction. Capacitive voltage multipliers exploit this idea.
So, the unique property of this circuit trick is that it can produce voltage outside the range limited by the supply rails - below the negative rail and above the positive rail. It does it by connecting the charged capacitor in series to the voltage source.
Of course, the capacitor gradually discharges and needs to be recharged from time to time (like the refresh in SRAM). That is why, I have called above this trick "dynamic". So, it can be applied in AC circuits.