How many ways are there of creating an $8$ character password with a digit, a lowercase letter, and $2$ capital letters?

There is no contradiction, just note that $C\cap D=\emptyset$ and therefore all intersection sets including $C\cap D$ have zero cardinality. Moreover, by a similar reason, also $A\cap B\cap C=\emptyset$ and $A\cap B\cap D=\emptyset$. Thus it remains to compute $$\begin{align} &|U|-(|A|+|B|+|C|+|D|)+(|A\cap B|+|A\cap C|+|A\cap D|+|B\cap C|+|B\cap D|) \end{align} $$ P.S. One more thing, your number for $|D|$ should be multiplied by $8$, i.e. the possible positions of the capital letter.

First, your evaluation of $|D|$ is incorrect. This is the number of ways to choose a capital letter for the first character, and then choose the remaining seven characters with no capitals. But in fact the single capital letter could be in any of $8$ positions, so actually $|D|=8\times 26\times 36^7$.

Secondly, the multiple events. Note that $C$ and $D$ cannot both occur, and no three events can occur (since if they include $A,B$ and one of $C,D$ then you have at most one character). So you only need to subtract the pairs of events (not including $|C\cup D|$), since the other terms are all $0$.

$|A\cap B|=26^8$, since this means all characters are capitals, and the other double events can be calculated similarly (ones involving $D$ are a bit more complicated, but no more so than the calculation for $D$ itself).