Let: $$S=\{m\alpha+n;\,m\in \mathbb N, n\in \mathbb Z\}$$ We claim that $\overline S =\mathbb R$. Then, $$S\subset K=\{a+b\alpha;a,b \in \mathbb Z\}$$ imply that $\mathbb R=\overline S\subseteq \overline K\subseteq \mathbb R$.
It's easy to see that, if $\zeta \in S$, then $k\zeta +m \in S$ for any $k\in \mathbb N$ and $m\in \mathbb Z$.
lemma: For every $m\in \mathbb N$, there is $\zeta \in S$ such that $|\zeta|\lt \frac 1m$.
Proof: Consider: $\zeta_n=n\alpha-[n\alpha]$ for $n=1,2,...,m+1$. ([x] denotes the greatest
integer not greater than x.) Then, $\zeta_n\in S$ and $0\lt \zeta_n \lt 1$ and more, if $k\neq j$, $0\neq \zeta_k-\zeta_j \in S$. Consider the $m$ intervals: $(\frac{j-1}{m},\frac{j}{m})$ for $j=1,2,...,m$. Each $\zeta_n$ is irrational and $0\lt \zeta_n \lt 1$. So, each $\zeta_n$, belong to one of this intervals. But, the number of $\zeta_n$ is $m+1$ and the number of this intervals, is $m$. Thus, there are $k\lt i$ such that $\zeta_k,\zeta_i \in (\frac{j-1}{m},\frac{j}{m})$ for some $1\le j\le m$. This means, $|\zeta_k-\zeta_i|\lt \frac 1m$ and $\zeta=\zeta_i-\zeta_k \in S$.
Now, let $x\in \mathbb R$ be arbitrary and $\epsilon \gt 0$ be given. We know that there is $m\in \mathbb N$, such that: $\frac 1m \lt \epsilon$. Now, by above lemma, there is $\zeta \in S$, such that: $|\zeta|\lt \frac 1m$. If $0\lt \zeta \lt \frac 1m$, we can find $n\in \mathbb Z$, such that, $x+n\gt 0$. then let $k\in \mathbb Z$ be the smallest positive integer such that: $k\zeta \gt x+n$. Then, $\eta=k\zeta-n\in S$ and: $x\lt \eta \lt x+\frac 1m$. If: $-\frac 1m \lt \zeta \lt 0$, we can chose $n\in \mathbb Z$, such that: $x+n \lt0$. Now, chose $k$ the largest positive integer such that: $k\zeta \lt x+n$ and put: $\eta=k\zeta -n \in S$. Then: $x-\frac 1m \lt \eta \lt x$.
