Trouble with Controlling LED with an Interrupt

Hello, everyone! I am new to both the forum, and Arduino itself, so please understand if this is posted in the wrong section, or if the error in my programming is obvious. What I am attempting to do with this sketch is light an LED when a push button, which acts as an interrupt, is switched on. I have wired everything correctly (LED wired to a digital pin, button wired to an interrupt pin), so I believe the error lies in the sketch itself. I added some lines for the board to print when the LED is supposed to light and when the interrupt is triggered; however, I do not see these lines when viewing the serial monitor, nor do I see the LED light up. Here is my code:

const int ledPin = 4;

volatile boolean ledState = false;
long ledInt;
long lastDebounceTime = 0;
long debounceDelay = 200;


void setup() {
  pinMode(ledPin, OUTPUT);
  attachInterrupt(4, ledOn, FALLING);
}

void loop() {
  debounce();

}
void debounce() {
  long lastDebounceTime = millis();
  if ( ledInt - lastDebounceTime > debounceDelay) {
    Serial.println(ledState);
    ledState = !ledState;
    lastDebounceTime = ledInt;
    buttonPress();
  }
}
void ledOn() {
  ledInt = millis();
}


void buttonPress() {
  if (ledState == true) {
    float something = millis() / 4000.0;
    int value = 128.0 + 128 * sin( something * 2.0 * PI  );
    digitalWrite(ledPin, value);
    Serial.println("LED on");
  }
  if (ledState == false) {
    digitalWrite(ledPin, LOW);
    Serial.println("LED off");
  }
}

Any help would be greatly appreciated--thank you!

Pin 4 isn't an external interrupt pin, only 2 and 3 (interrupts 0 and 1)

Edit: I'm confused, where is the LED wired and where is the switch wired?

AWOL:
Pin 4 isn't an external interrupt pin, only 2 and 3 (interrupts 0 and 1)

Edit: I'm confused, where is the LED wired and where is the switch wired?

The LED is wired to pin 4, and the button is wired to pin 7, which is read by the Arduino Leonardo (the board I am using) as interrupt 4. Sorry for the confusion :3

What I am attempting to do with this sketch is light an LED when a push button, which acts as an interrupt, is switched on.

Why? Turning an LED on or off immediately (within nanoseconds) when a ssslllooowww human presses a switch is pointless.

Why are you doing Serial.print() in the interrupt service routine?

I am doing Serial.println as a form of debugging, so I can see when that part of the code is active.

Don't do serial prints in an ISR. They can lock up the processor. Now you have two problems. The original one and the one you introduced with your debugging print.