Somehow I get the feeling you could do with some basics...

DIGITAL only has two states, On or Off, High or Low, 1 or 0. That's why we call it Binary.

A PIC pin when used for DIGITAL can only have two states... High (or 1) or Low (0). 1 or Zero that's all you got. There's no 'in-between', no half-way or part way. Your choice is 1 or Zero. That's it - no other choices. This is also why everybody does DIGITAL and very few people do ANALOG. Because with DIGITAL life is easy. You learn 1 and you learn Zero and that's it - you're a computer expert because there's nothing else to learn.

This means one of two things...

Scenario A. You preset the PIC's pin with a HIGH (ie set it to a Logic 1) and you do this by pulling it UP to +5v via a Resistor (say 10K for arguments sake). This pin is normally now sitting at +5v (Logic 1 or High). You connect a pushbutton or switch between the PIC's pin and Ground (0v). When you push the button, you short that pin down to 0v, forcing it LOW. You need a Resistor, because otherwise you'll be shorting the +5v Power Supply out! Lots of PICs have pull-up Resistors inside them on many of their pins which can be switched-in to do the job and save you from providing one externally. So here, with the switch being OPEN-circuit, the PIC sees a HIGH (Logic 1) on it's pin, and when you press the Button the PIC sees a LOW (0) on it's pin.

Scenario B. You preset the PIC's pin with a LOW (ie set it to a Logic 0) and you do this by pulling it DOWN to 0v via a Resistor (say 10K for arguments sake). This pin is normally now sitting at 0v (Logic 0 or Low). You connect a pushbutton or switch between the PIC's pin and Supply (+5v). When you push the button, you connect that pin up to +5v, forcing it HIGH. You need a Resistor, because otherwise you'll be shorting the +5v Power Supply out! PICs don't carry internal pull-down Resistors inside, so you need to provide one externally. So here, with the switch being OPEN-circuit, the PIC sees a LOW (Logic 0) on it's pin, and when you press the Button the PIC sees a HIGH (1) on it's pin.

You CANNOT leave a PIC pin 'floating' unconnected. This is because the pin can acquire a charge (or discharge) and set the pin HIGH or LOW or even flicker between the two - which is kinda defeating what you want the pin to do if you can't even predict what state it is going to be in at any point in time. You pull it LOW with a switch and it might stay there when the switch is released, then some time later it might float High. You pull it High with a switch, then it might stay there forever after you release the switch - again not what you want it to do.

And in answer to your question - No, you won't damage a PIC's pin (when it's set for INPUT - TRISx=1) by connecting it directly to +5v or to 0v.