Exercise 6: Duplicate Letters — the Shorter Pythonic Way — Possible Solution
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word = "bookkeeper"

duplicates = []

for ch in word:
    if ch not in duplicates and word.count(ch) > 1:
        duplicates.append(ch)

print(duplicates)

Output:
['o', 'k', 'e']

WHY THIS WORKS AS AN ANSWER
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Same result as Exercise 5's two-list version, and that's the whole
point — two genuinely different approaches to the same problem
should always agree, or one of them has a bug. Here, word.count(ch)
> 1 answers "does this character appear more than once anywhere in
the word?" directly, without needing a running my_list at all. The
ch not in duplicates guard is still doing the same job the elif did
in Exercise 5 — stopping "e" from being appended three separate
times just because word.count("e") > 1 is True on every one of its
three occurrences.

Worth noticing: this version is shorter, but per this lesson's own
finding-box, word.count(ch) rescans the entire 10-character word on
every single pass through the loop — 10 rescans total for a
10-character word. For "bookkeeper" that's completely invisible; on
a long paragraph, it stops being invisible.