Breaking out of JavaScript 'For' Loop using False?

The return statement ends the function , so the return true line is unreachable. Hope that helps.

The below code apparently logs values 1 to 5, then breaks out of the 'for' loop, because the 'false' value is returned.

Wrong, it breaks out of the for loop because of the return, not because of the false. You could write return x with the same effect. Point is, return immediatly drops out of the enclosing function, whatever loop or conditional is being executed currently.

The loop will never be executed till 10. return false will return control to caller of the function x as soon as i is equal to 5. The return true line is unreachable.
This is your code

function x() {
    for (var i = 0; i < 10; i++) {
        console.log(i);
        if (i == 5) return false;
    }
    return true
}

console.log(x());

Output:-

 0
 1
 2
 3
 4
 5
 false

If you want it reach to the return true statement you may want to do something like adding a break instead of return false in for loop.

for (var i = 0; i < 10; i++) {
            console.log(i);
            if (i == 5) break;//this will put it outside for loop
        }

If you do not want mess with return or break statements, you can add just one boolean flag. In my example i use isFound flag to run loop until your requirements are meet or end of array.

const arr = [1,2,3,4,5];

for(let i = 0, isFound; !isFound; i++){
  if(i >= 3 || i >= arr.length){isFound = true;}
  else { console.log(arr[i]);}
}

So the output will be:

1
2 
3