Let $\Sigma$ be the Borel $\sigma$-algebra of $\mathbb{R}$. We define a measure $\mu$ on $\Sigma$. For every Borel set $A$ we define $\mu(A)$ to be the number of elements from the set $\{\frac{1}{n}: n\in\mathbb{N}\}$ which are in $A$. The question asks to prove this measure is not regular. From what I know $\mu$ is called regular if for any measurable set $A$ the following holds:
$\mu(A)$=sup{$\mu(K)$: $K\subseteq A$, $K$ is compact and measurable}
The problem is that I got the impression that the statement I need to prove is simply wrong. Let $A\in\Sigma$. If $\mu(A)<\infty$ then we can take $K=A\cap\{\frac{1}{n}:n\in\mathbb{N}\}$. This is a compact subset of $A$ which has exactly the same measure as $A$. If $\mu(A)=\infty$ then we can write $A\cap\{\frac{1}{n}:n\in\mathbb{N}\}=\{x_m\}_{m=1}^\infty$ using the fact that this is a countable set. And then for every $m\in\mathbb{N}$ we define $K_m=\{x_1,x_2,...,x_m\}$. We get that $K_m$ is a measurable compact subset of $A$ which has measure $m$. Because we define that for all $m\in\mathbb{N}$ we get that for any $M>0$ there is a measurable compact subset $K\subseteq A$ with $\mu(K)>M$. Hence the supremum of the measures of the compact subsets of $A$ is $\infty$, just like $\mu(A)$. So $\mu$ is actually regular.
Am I missing something here, or is the statement really wrong?
Your proof is correct.
But if we use the definition from Wikipedia, which has the additional condition $$ \mu(A)=\inf\{\mu(O): O\supseteq A, O\;\text{is open and measurable}\}. $$ It turns out that the measure $\mu$ does not satisfy this condition (leave a comment if you need help to understand why this is the case).
(Note: The definition of a regular measure probably varies in the literature. Even in wikipedia an alternative version is mentioned. You have to check your source material which definition of regular measure is used.)