For the learners: you should know that doing something like the setup for this challenge inclines you to do is a bad practice. Never do the following:

def f():
	if condition:
    	return True
    else:
    	return False

This is just dumb. You are returning a boolean, so why even use if blocks in the first place? The correct what of doing this would be:

def f():
	return condition

Because this already evaluates as a boolean.

So in this challenge, forget about ifs and elses, and that leap variable, and just do the following:

def is_leap(year):
    return year % 4 == 0 and (year % 400 == 0 or year % 100 != 0)

Don't be redundant, be DRY.

The calendar builtin module has a function isleap() that returns True if the year is leap, otherwise it returns False

import calendar

def is_leap(year):
    return calendar.isleap(year)

The problem's explanation is not clear. I am not an english speaker and the bullet points were not clear enough for me.

Can that be improved?

Suggested explanation is bellow:

In the Gregorian calendar three criteria must be taken into account to identify leap years

The year can be evenly divided by 4;

If the year can be evenly divided by 100, then it is not a leap year

The year is also evenly divisible by 400 then it is a leap year.

Working

if year%4 == 0 and (year%100 != 0 or year%400 == 0): leap = True

Method 1

def is_leap(year):
	return year % 4 == 0 and (year % 400 == 0 or year % 100 != 0)

Method 2

def is_leap(y):
    return (y%400==0) or (y%100!=0 and y%4==0)

Method 3

def is_leap(year):
    leap = False
    
    if (year % 4 == 0):
        leap = True
        if (year % 100 == 0):
            leap = False
            if (year % 400 == 0):
                leap = True
    return leap