There exists $R>r$ such that $\overline{D}(a,r)\subset {D}(a,R)\subset U$ Let $U$ be an open set of $\mathbb{C}$ and $a\in U$ and $r>0$ we suppose that $$\overline{D}(a,r):\{z\in \mathbb{C}:|z-a|\leq r\}\subset U$$
can we say that there exists $R>r$ such  that  $$\overline{D}(a,r)\subset {D}(a,R)\subset U$$
with ${D}(a,R):\{z\in \mathbb{C}:|z-a|< R\}$
An idea please?

If such a $R$ didn't exist, then for all $R>r$, there would exist $z\in D(a,R)$ such that $z\notin U$. Taking $R=r+\frac{1}{n}$, you can construct $(z_n)_{n\geqslant 1}$ such that $|z_n-a|\leqslant r+\frac{1}{n}$ and $z_n\notin U$ for all $n$. In particular $z_n\in\overline{D}(a,r+1)$ and since $\overline{D}(a,r+1)$ is compact, there exists $\sigma:\mathbb{N}\rightarrow\mathbb{N}$ and $z_{\infty}\in\overline{D}(a,r+1)$ such that $\lim\limits_{n\rightarrow +\infty}z_{\sigma(n)}=z_{\infty}$. Since $\mathbb{C}\setminus U$ is closed, $z_{\infty}\notin U$, on the other hand $|z_{\sigma(n)}-a|\leqslant r+\frac{1}{\sigma(n)}$ for all $n\geqslant 1$, taking the limit as $n\rightarrow +\infty$ gives that $z_{\infty}\in\overline{D}(a,r)$, this is absurd as $\overline{D}(a,r)\subset U$.
