It is true that your 2nd approach, as it is stated in the posted question, is erroneous. It is also true that the error is as stated in your answer: you are over-counting. However, your 2nd approach can be repaired through the use of Inclusion-Exclusion. See this article for an
introduction to Inclusion-Exclusion.
Then, see this answer for an explanation of and justification for the Inclusion-Exclusion formula.
Let $~S~$ denote the set of all possible $~9~$ digit numbers, without any regard for whether the number contains an $~8~$ anywhere in its digits. For $~k \in \{1,2,\cdots,9\},~$ let $~S_k~$ denote the subset of $~S~$ that specifically contains an $~8~$ in digit position 1 (where the digit positions read from left to right).
For example, an element in $~S_1~$ would be any number that contains an $~8~$ in digit position 1. Such an element may or may not contain an $~8~$ in some other digit position.
Then, the desired enumeration is
$$|S_1 \cup S_2 \cup \cdots \cup S_9|. \tag1 $$
Let $~T_1~$ denote $\displaystyle \sum_{1 \leq i_1 \leq 9} S_{i_1}.~$
That is, $~T_1~$ denotes the sum of $~\displaystyle \binom{9}{1}~$ terms.
By considerations of symmetry, you have that
$~|S_1| = |S_2| = \cdots = |S_9|.$
Therefore, $~T_1 = 9 \times |S_1| = 9 \times (10)^{(9-1)}.$
Similarly, for $~r \in \{2,3,\cdots,9\},~$
let $~T_r~$ denote $~\displaystyle \sum_{1 \leq i_1 < i_2 < \cdots < i_r \leq 9} | ~S_{i_1} \cap S_{i_2} \cap \cdots \cap S_{i_r} ~|.$
That is, $~T_r~$ denotes the sum of $~\displaystyle \binom{9}{r}~$ terms.
Then, by similar considerations of symmetry, you have that
$$T_r = \binom{9}{r} \times (10)^{(9-r)}.$$
By Inclusion-Theory, the computation in (1) above is equivalent to
$$\sum_{r=1}^9 (-1)^{(r+1)} T_r$$
$$= \sum_{r=1}^9 (-1)^{(r+1)} \left[ ~\binom{9}{r} \times (10)^{9-r}~\right]. \tag2 $$
The expression in (2) above is in fact equivalent to $~(10)^9 - 9^9.$
This can be demonstrated by taking the binomial expansion of
$~9^9 = (10 - 1)^9.$
$$9^9 = (10 - 1)^9 = \sum_{r = 0}^9 \left[ ~\binom{9}{r} (10)^{(9-r)} (-1)^r ~\right] \implies $$
$$10^9 - 9^9 = \sum_{r = 1}^9 \left[ ~\binom{9}{r} (10)^{(9-r)} (-1)^{(r + 1)} ~\right].$$
$~\underline{\text{Addendum}}$
Still another approach is to reason that if a 9 digit number has at least one 8, then it must have exactly $~k~$ 8's, where $~k~$ is some element in $~\{1,2,\cdots,9\}.$
So, you have the mutually exclusive possibilities, based on the value of $~k,~$ where the enumeration of 9 digit numbers with exactly $~k~$ 8's is
$$\binom{9}{k} 9^{(9-k)}.$$
Therefore, you have an alternate computation of
$$\sum_{k=1}^9 \left[ ~\binom{9}{k} 9^{(9-k)} ~\right]. \tag3 $$
The expression in (3) above can also be demonstrated to be equivalent to $~(10)^9 - 9^9,~$ by taking the binomial expansion of
$~(10)^9 = (9 + 1)^9.$
$$(10)^9 = (9 + 1)^9 = \sum_{k = 0}^9 \left[ ~\binom{9}{k} 9^{(9-k)} 1^k ~\right] \implies $$
$$10^9 - 9^9 = \sum_{k = 1}^9 \left[ ~\binom{9}{k} 9^{(9-k)} ~\right].$$
