Amount of numbers $n \in \{1,\ldots,9999\}$ such that there aren't $2$ consecutive odd digits I want to find out the amount of numbers $n \in \{1,\ldots,9999\}$ such that there aren't $2$ consecutive odd digits. I want to use the principle of inclusion and exclusion.
There are 9999 numbers in $\{1,\ldots,9999\}$.
$n=ABCD$
$A_1:=\{n\in \mathbb N:n<10000,\ A,B \mathrm\ {odd}\}$
$A_2:=\{n\in \mathbb N:n<10000,\ B,C \mathrm\ {odd}\}$
$A_3:=\{n\in \mathbb N:n<10000,\ C,D \mathrm\ {odd}\}$

I don't really know how to deal with the fact that $n$ could have less than $4$ digits.

$|A_1|=5\cdot5\cdot10\cdot10$
$|A_2|=10\cdot5\cdot5\cdot10+5\cdot5\cdot10$ (the second summand because $n$ in $A_2$ could have $3$ digits instead of $4$, but I don't know if this is correct)
$|A_3|=10\cdot10\cdot5\cdot5+10\cdot5\cdot5+5\cdot5$ (4 digits, 3 digits or 2 digits)
$|A_1\cap A_2|=5\cdot5\cdot5\cdot10$
$|A_1\cap A_3|=5\cdot5\cdot5\cdot5$
$|A_2\cap A_3|=10\cdot5\cdot5\cdot5+5\cdot5\cdot5$
$|A_1\cap A_2\cap A_3|=5\cdot5\cdot5\cdot5$
Are my calculations correct?
 A: Thanks to N.F.Taussig for suggesting that my answer should be self-contained.  I have edited my answer.  Also, thanks to N.F. Taussig for indicating an analytical error that I made.  I have corrected the answer.

You can assume that the numbers run from $0$ through $9999$, to ease the calculations.  Then, at the end, you can adjust for $0$ being considered.
Further, since what is being enumerated is two consecutive odd digits, it is harmless to adopt the simplification that all numbers are zero filled on the left.

Re my two comments, following the question, this is how I would employ Inclusion-Exclusion:
My approach will be to enumerate all possible occurrences of at least two consecutive odd digits.  Then, I will deduct this from $(9999)$.
Edit
The foundation of Inclusion-Exclusion, is that by counting - deducting - adding back, each pertinent occurrence (in this case of having at least one pair of consecutive odd integers) ends up being counted once.
So, in this particular problem, I made a mistake in not analyzing closely, how many times various intersections are counted in my enumerations of $T_1, T_2, T_3$, as discussed below.  I have added discussion.
In the discussion below, I will use 
o-o-e-e 
(for example) to indicate that (reading from the left), the first two digits are odd and the second two digits are even.
Similarly, I will use 
o-o-o-e 
to indicate that the first three digits are odd and the last digit is event.

Number of ways first two digits odd (reading the digits from left to right):
$5^2 \times 10^2.$ 
Multiply the above by $3$, since you are also enumerating digits $2,3$ odd and digits $3,4$ odd. 
$T_1 = 3 \times 5^2 \times 10^2 = 7500.$
In determining how many collections of $4$ digits to deduct, re intersections, consider the following intersections, which are inherent in my enumeration of $T_1$:
o-o-e-e $~:~$ counted once. 
e-o-o-e $~:~$ counted once. 
e-e-o-o $~:~$ counted once. 
o-o-o-e $~:~$ counted twice. 
e-o-o-o $~:~$ counted twice. 
o-o-o-o $~:~$ counted three times. 
This was my oversight.

Deduct number of ways that digits $1,2,3$ are odd, or that digits $2,3,4$ are odd: 
$T_2 = 2 \times 5^3 \times 10 = 2500.$ 
Edit
Re my oversight, I must also consider having at least two pairs of odd integers, where the pairs are represented by $(1,2)$ and $(3,4)$.
So, the corrected computation of $T_2$ is: 
$T_2 = \left[2 \times 5^3 \times 10\right] + 5^4 = 3125.$ 
o-o-o-e $~:~$ deducted once. 
e-o-o-o $~:~$ deducted once. 
o-o-o-o $~:~$ deducted three times. 
At this point, everything is all square, with all pertinent intersections having been counted exactly once, except that (now), o-o-o-o has been counted three times and deducted three times.

Add back number of ways that all 4 digits are odd. 
$T_3 = 5^4 = 625.$
Final computation:
$$T_1 - T_2 + T_3 = 7500 - 3125 + 625 = 5000.$$
Edit
Confess: after reading N.F. Taussig's comments following my answer, I sanity checked with a Java program.

Since you are actually interested in computation where there aren't two consecutive odd digits, you actually want:
$$9999 - 5000 = 4999.$$
A: An answer of almost half the numbers appears suspicious to me, so here is a direct count using another method:
If we consider the number string $0000\;\; thru\;\; 9999$, and finally exclude $0000,$ there are eight types of number strings that satisfy the criteria, viz
$\displaylines{o-e-e-e\\ o-e-o-e\\o-e-e-o\\e-o-e-e\\ e-e-o-e\\e-e-e-o\\ e-o-e-o\\ e-e-e-e}$
Thus total valid numbers  are $8\cdot5^4 - 1 = 4999$
which confirms the answers using inclusion-exclusion
