Are internal pull-up resistors in microcontrollers connected to Vcc internally?
The internal pull-ups (and pull-downs, if they also exist) are usually relatively weak and are made from PMOS (or NMOS) devices. They usually can support something on the order of about \$30\:\mu\text{A}\$.

When you enable one of these, the software action simply applies an appropriate gate voltage to the device so that it allows a small current to flow. For a pull-up, one side of the device is tied to the \$V_{CC}\$ you also provide via a separate pin to the device. For a pull-down (if available), one side of the device is tied to the ground you also provide via a separate pin. Either way, the other side of the device is tied to the I/O pin.

You can use the internal pull-up to provide a weak current source for use with an external switch. So it is possible (and not infrequently done) to use it with an external switch without having to add an external resistor.

However, this internal pull-up is usually weaker than an externally added resistor (\$\approx 100\:\text{k}\Omega\$ for the internal pull-up versus a not-uncommon \$10\:\text{k}\Omega\$ used when adding an external resistor.) For a simple push-button that is kept close to the MCU, it may not matter so much. But if the switch is placed far away or is used within a rather noisy environment, you could be in trouble using the internal pull-up. Also, if you use the I/O pin for both IN and OUT (muxing it for some reason) then again there may be a reason for something placed externally, instead.

In your case, with an internal pull-up, you can simply hook one side of the switch to the I/O pin and the other side to ground.