Your code looks well-formatted, it's easy to read and to follow, and the explanation you gave matches the code exactly. Well done. :)
for item in my_list:
This statement looks strange since in the body of this for loop, you neither use item nor my_list. You can express the idea of that code more directly:
for _ in range(len(my_list)):
The variable _ is just a normal variable, and you could even access its value. By convention though, the variable _ means an unused variable.
Or, alternatively, you could express it even shorter:
while len(unsorted_list) > 0:
This wording matches the body of the loop, where you remove an element from unsorted_list in each iteration. This is good. It's still quite long though, therefore the idiomatic way to test whether a list is empty is:
while unsorted_list:
This makes your code:
def my_sort(my_list):
unsorted_list = [*my_list]
sorted_list = []
while unsorted_list:
shortest_num = inf
for num in unsorted_list:
if num < shortest_num:
shortest_num = num
unsorted_list.remove(shortest_num)
sorted_list.append(shortest_num)
return sorted_list
This code can be made much shorter since Python has a built-in function called min. By using it you can transform your code to:
def my_sort(my_list):
unsorted_list = [*my_list]
sorted_list = []
while unsorted_list:
shortest_num = min(unsorted_list)
unsorted_list.remove(shortest_num)
sorted_list.append(shortest_num)
return sorted_list
This is already much shorter than before, and you don't need to import math.inf anymore, which is good, since it was a placeholder anyway that was not really necessary from a high-level point of view.
This is about as elegant as it gets. You should not use this sorting algorithm for sorting large data since for a list of \$n\$ elements it requires \$ n + n \cdot (\frac n2 + \frac n2 + 1) \$ steps, or a bit more compressed, it has the time complexity \$\mathcal O(n^2)\$. There are faster sorting algorithms out there. These are more complicated to implement, but once they are implemented and tested, they are as easy to use as your algorithm.