Is every prime contained in the set $\,\{n\in\mathbb{N}:n|10^k-1\}\cup\{2,5\}\,$ for $k=1,2,3,...$? 
Is every prime contained in the set
  $\,\{n\in\mathbb{N}:n|10^k-1\}\cup\{2,5\}\,$ for $k=1,2,3,...$?

Let $p\notin\{2,5\}$ be a prime number. Then $\frac{1}{p}$ has a decimal period of length at most $p-1$. Denote $\frac{1}{p}$ by $$\frac{1}{p}=0.\overline{d_1d_2...d_r}$$ where $1\leq r \lt p$ and $d_i\in\{0,1,2,...,9\}$.
Let $d=0.d_1d_2...d_r\cdot10^r\;$ (e.g., if $p=7$, then $\frac{1}{7}=0.\overline{142857}$, $r=6$, and $d=142857$).
Let's consider $\frac{1}{10^m-1}$ for some $m\in\mathbb{N}$:
$$
\frac{1}{10^m-1}=\frac{1}{\underbrace{99...9}_{\text{$m$ times}}}=0.\overline{\underbrace{00...0}_{m-1 \\ \text{times}}1}=\sum_{i=1}^\infty \frac{1}{10^{i\cdot m}}
$$
So
\begin{align}
\frac{1}{p}&=0.\overline{d_1d_2...d_r}=d\cdot \left(\frac{1}{10^r}\right)+d\cdot \left(\frac{1}{10^{2r}}\right)+d\cdot \left(\frac{1}{10^{3r}}\right)+\cdots \\ &=d\cdot\left(\frac{1}{10^r}+\frac{1}{10^{2r}}+\frac{1}{10^{3r}}+\cdots\right) \\ &=d\cdot\sum_{i=1}^\infty \frac{1}{10^{i\cdot r}} \\ &=\frac{d}{10^r-1}
\end{align}
Rearranging terms, we have $\,10^r-1=p\cdot d \, \Longrightarrow \, p|10^r-1$.
Does this prove that the set of primes is contained in the set $\{n\in\mathbb{N}:n|10^k-1\}\cup\{2,5\}$?
Clearly, $2$ and $5$ are missing from $\{n\in\mathbb{N}:n|10^k-1\}$. Are there any other primes not contained in this set?

Yes: it's true that every prime $p \neq 2$ or $5$ divides some number of the form $10^k-1$.
There is a simple way to prove this fact.
Look at the remainders of $10^k-1$ when divided by $p$. Clearly there are only finitely many remainders, namely $0,1,2, \dots, p-1$. Since the numbers of the form $10^k-1$ are infinitely many, there are two of them which give the same remainder when divided by $p$.
Let's say there are $k<l$ such that $10^k-1$ and $10^l-1$ have the same remainder when divided by $p$. This is equivalent on saying that
$$(10^l-1)-(10^k-1)$$
is a multiple of $p$. But
$$(10^l-1)-(10^k-1) = 10^l-10^k = 10^k (10^{l-k}-1)$$
Since $p$ divides $10^k (10^{l-k}-1)$ and does not divide $10^k$, necessarily $p$ divides $(10^{l-k}-1)$. This concludes the proof.
