Do all 'or' cases need to be checked in a direct proof? There's a problem from Richard Hammack's book of proof, which asks to prove the proposition: 
Suppose $x,y \in \mathbb{Z}$. If $5 \nmid xy$, then $5 \nmid x$ and $5 \nmid y$
In this case, the intent is to show the use of the contrapositive proof. So a direct proof on $$5\mid x \lor 5\mid y \implies 5 \mid xy$$
The proof is straightforward, and in the book the solution shows that when $5\mid x$ is true, then $5 \mid xy$ is true, and similarly for $5\mid y$. 
In this case, the solution doesn't show the case where both $5\mid x$ and $5\mid y$ are true. 
In this particular example the proof of the third case is trivial, and so I understand that it can be left out entirely.
However, I am wondering in general, when using direct proof, and when the $P(x,y)$ statement is compound statement made up of two statements conjuncted with an or, then is it necessary to check all 3 cases? 
In other words, to link it back to the example proof, is it strictly necessary to show that $5\mid x$ AND $5\mid y$ to complete the proof? Or is it that since we have shown the T/F and F/T case, the T/T case is also always true?
 A: The case $5|x$ and $5|y$ is already covered by the other cases, so it is not necessary to consider it separately. The important thing is to make sure that the cases you consider are exhaustive. 
A: Unpacking the book's reasoning a bit, 

when $5\mid x$ is true, then $5\mid xy$ is true

can be expanded to

$5\mid x$ implies $5\mid xy$, and since I did not say anything about $5\mid y$ just then, the implication is true when $5\mid y$ is true and it also is true when $5\mid y$ is false.

Or to put it another way, when someone says "suppose $5\mid x$ is true," it is faulty logic to assume that means $5\mid y$ is false. If you don't prove that $5\mid y$ is false, or make it an explicit additional assumption, then you can't say anything that depends on $5\mid y$ being false, nor can you say anything that depends on $5\mid y$ being true.
In order to make a proof of the statement "if $5\mid x$ is true then $5\mid xy$ is true," you must use reasons that are valid no matter what the truth status of $5\mid y$ is. I guess there are details in the book that you did not include in the question; presumably they obey those requirements.
A: To rephrase @Mauro ALLEGRANZA's comment: What you have proven is NOT the 'T/F' and 'F/T' cases, they are instead the 'T/?' and '?/T' cases, where by '?' I mean you don't know that part's truth value.  Since you are in each case OR-ing 'T' and '?', you always get 'T' regardless of which one '?' is.
A: Consider it this way: If you have separately proven that $5|x$ is true and $5|y$ is true, you can linguistically say that you have proven that the statements $5|x$ and $5|y$ are both true. This is logically the same as $5|x \wedge 5|y$ being true, so you do NOT need to separately check this, since you have already done so logically. Typically, the "and" statement is not considered a separate case, so a proof by cases comprises only of proving the separate statements. 
