I've tested this and it works but it doesn't pass the challenge

While I personally encourage you to test in an external REPL as it can help you test and understand your code before you submit it, I do have to say that Steven's example exposes your problem fairly well. He's right that no word appears more than twice. However, when I test your function with his example, this is the output I get:

>>> word_count("What is love Baby don't hurt me Don't hurt me no more")
{'what': 1, 'is': 1, 'love': 1, 'baby': 1, "don't": 2, 'hurt': 3, 'me': 4, 'no': 1, 'more': 1}

You'll notice that 'hurt' is 3 and 'me' is 4. That's not correct, because as Steven and I said before, no word appears more than twice. In your code, the only place that a number is set in word_dictionary is from your variable count, so this means that your variable count is getting out of sync with reality.

In this case, it's a scoping issue. Before the loop starts, you set count to 1. It's important to note that this was declared before the loop began, and thus it exists and persists with the same value for each iteration of the loop without being reset by Python. In your code, when the loop encounters a word that doesn't exist as a key in the word_dictionary yet, it resets count to 1, which is fine. However , because count holds its value between iterations, this can be a problem when the loop encounters a few words in a row that are already in word_dictionary.

Let's jump ahead to the 8th iteration of the loop. Right now, the word we're on from the original sentence is "Don't", but it's been lowercased earlier in the function. This is what word_dictionary looks like:

{'what': 1, 'is': 1, 'love': 1, 'baby': 1, "don't": 1, 'hurt': 1, 'me': 1}

Up to this point, every word we've encountered has been unique, so the count variable has been reset to 1 every time. However, now that we're on "don't", we've seen this word before, so we go onto the else block. 1 is added to count, so count is now 2. This is a correct final value, but as you'll see in a moment, that's coincidental. word_dictionary["don't"] is set to count (which, again, is 2 at this point), and we continue onto the next iteration of the loop

At this iteration, the word we're on is the 2nd "hurt" in the sentence. Since it's outside the loop, count has hung onto its value that it's had before, which is still 2 at this point. We've seen this word before, so we go onto the else block again. We add 1 to count so that it's 3 now, and we assign that to word_dictionary["hurt"].

The same thing happens to "me" on the next iteration. count is 3 when the loop starts, and when the else block runs, count is incremented once more, so word_dictionary["me"] is set to 4.

I know I've been stressing throughout this answer that the fact that count is declared outside the loop, but this was just to illustrate why it goes too high on certain words. Ultimately, placing it inside the loop isn't a very good solution, either, because then its value will always be either 1 or 2 depending on if the word was encountered before, and won't ever go higher, even if the word is found 3 or more times. Instead of holding onto an external count variable, I recommend checking the value in the dictionary on each iteration of the loop and setting it to 1 if the key doesn't exist yet or increment it if it does