$C_1+C_2$ not closed with $C_1;C_2 \subset \mathbb{R}$ closed
I thought of $C_1=\mathbb{Z}^{-}; C_2=\{n + \frac{1}{n} \mid n \in \mathbb{Z^+}; n \geq 3\}$
$C_2$ is closed as
$$|k+\frac{1}{k}-(l+\frac{1}{l})|= |k-l-(\frac{1}{l}-\frac{1}{k})| \geq |k-l| -|\frac{1}{l}-\frac{1}{k}| \geq |k-l|-\frac{2}{3}; \quad k;l\geq3$$ So if $b \in cl(C_2)$, then
$$B\big(b,\frac{1}{6}\big) \cap X \neq \emptyset$$
Let $x \in B\big(b,\frac{1}{3}\big) \cap X$, then as shown above $|x-y| \geq \frac{1}{3}$ for $y \in X$, so there is no other point $p$ in $X \cap B(b,\frac{1}{6})$ and so, we should have that $b=x$.
But $-n+(n+\frac{1}{n})=\frac{1}{n} \in C_1 + C_2 \; (n \geq 3)$and $0 \notin C_1 + C_2$ as if that were the case, we'd have $k+n=-\frac{1}{n}$ but $-\frac{1}{n}$ is not an integer as $n \geq 3$.
Is this a valid example? Any other suggestions?
