uint8_t elements exceeding one byte?

I saw this code:

uint8_t broadcastAddress[] = {0xFF, 0xFF, 0xFF, 0xFF, 0xFF, 0xFF};

I know uint8_t holds a byte, so how does this line of code work when a single element, 0xFF, is in itself one whole byte and the array is therefore 6 bytes wide.
Can someone explain how 6 bytes of data are of type single byte?

To try and probe it, I wrote this code on an Arduino Uno R3:

void setup() {
    Serial.begin(115200);
    // #include <stdint.h>  // Not needed in Arduino IDE with Arduino board selected.
    // #include <stdio.h>  // Not needed in Arduino IDE with Arduino board selected.
}

void loop() {
    delay(3000);
    uint8_t numbers[] = {0xFF, 0x22, 0xA0, 0x3B, 0xC2};  // 5 elements & 5 bytes

    Serial.print("sizeof(numbers) ");
    Serial.println(sizeof(numbers));   // outputs = 5, (obviously the # elements in array)

    for (int i = 0; i <= 8; i++) {
        Serial.print("sizeof(numbers[");
        Serial.print(i);
        Serial.print("]: ");
        Serial.println(sizeof(numbers[i]));  // outputs 1 for i=0 to 8! Element value vanished! non-used elements positions equate to same as used element positions 
    }
}

Solution

saw this code: uint8_t broadcastAddress[] = {0xFF, 0xFF, 0xFF, 0xFF, 0xFF, 0xFF};

I know uint8_t holds a byte, so how does this line of code work when a single element, 0xFF, is in itself one whole byte and the array is therefore 6 bytes wide.

broadcastAddress is not a uint8_t, but rather an array of uint8_ts ([] makes it an array).
It can therefore contain multiple elements of type uint8_t.
The compiler will determine the actual size of the array based on the initializer ({0xFF, 0xFF, 0xFF, 0xFF, 0xFF, 0xFF}) - i.e. 6 in this case.

The same applies to numbers in your test code (with 5 elements).