How can we show that $\left\{x\in\mathbb R^d:d(x,\Lambda^c)>\varepsilon\right\}\subseteq\Lambda$, if $\Lambda\subseteq\mathbb R^d$ is a domain?

Question

Let $d\in\mathbb N$ and $\Lambda\subseteq\mathbb R^d$ be connected and open. Now, let $\varepsilon>0$ and $$\Lambda_\varepsilon:=\left\{x\in\mathbb R^d:d(x,\Lambda^c)>\varepsilon\right\},$$ where $d$ denotes the Euclidean metric on $\mathbb R^d$. It's a simple question, but for the moment I don't see how we can show that $\Lambda_\varepsilon\subseteq\Lambda$.

Answer 1

Let $x \in \Lambda_{\epsilon}$. Suppose that on the contrary, $x \not \in \Lambda$. Therefore, $x \in \Lambda^c$ and in particular $d(x, \Lambda^c) = 0$, which is absurd because we're supposing that $d(x,\Lambda^c)$ is positive.

Edit: note that here we're not using anything about $\Lambda$, in general if $(X,d)$ is a metric space, $\Lambda \subseteq X$ and $\Lambda_{\epsilon} = \{x \in X : d(x,\Lambda^c) > \varepsilon \}$, it holds that $\Lambda_{\epsilon} \subseteq \Lambda$.