# Equivalence of two notions of separation in a metric space

I want to show that that the following two notions of separation in a metric space are equivalent

$$\textbf{1st Definition:}$$ Let $$(X,d)$$ be a metric space then a subset $$E$$ of $$X$$ is said to be separated if there exists two non-empty and disjoint open sets $$U_1$$ and $$U_2$$ in $$X$$ such that $$E \subset U_1 \cup U_2$$ and $$E\cap U_1 \neq \emptyset \neq E \cap U_2$$.

$$\textbf{2nd Definition:}$$ Let $$(X,d)$$ be a metric space then a set subset $$E$$ of $$X$$ is said to be separated if there exists non-empty and disjoint open sets $$V_1$$ and $$V_2$$ in metric space $$(E,d\restriction _{E\times E})$$ such that $$E = V_1 \cup V_2$$.

$$\textbf{My attempt:}$$ It's pretty easy to show that $$(1st) \implies (2nd)$$ but I am stuck at the reverse implication. If a subset $$E$$ of $$X$$ is separated according to 2nd defnition then $$E = V_1 \cup V_2$$ for some non-empty and disjoint open sets $$V_1$$ and $$V_2$$ in $$E$$. Therefore there exists open sets $$U_1$$ and $$U_2$$ in the ambient space $$X$$ such that $$V_1 = E \cap U_1$$ and $$V_2 = E \cap U_2$$. $$U_1$$ and $$U_2$$ satisfies all the conditions of 1st definition except disjointness. If I take $$U_3 = U_1 \cap (\overline{U_2})^{c}$$ to replace $$U_1$$ in order to impose disjointess I find it difficult to prove that $$V_1 = E \cap U_3$$.

First note that you cannot take arbitrary $$U_1$$ and $$U_2$$, for example $$U_2 = X$$ won't be good for your argument.
We can construct $$U_1$$ and $$U_2$$ so that they are disjoint. For every $$x\in V_1$$ consider $$r_x = d(x,V_2)/3$$. Obviously, $$B(x,r_x)\cap V_2 = \emptyset$$ (balls are in $$X$$). We take $$U_1 = \bigcup\limits_{x\in V_1} B(x,r_x)$$, $$U_2$$ is constructed similarly. We will show that they indeed are disjoint.
Assume that there exists $$z\in U_1\cap U_2$$. Then, by definition, $$z\in B(x,d(x,V_2)/3)$$ for some $$x\in V_1$$ and $$z\in B(y,d(y,V_1)/3)$$ for some $$y\in V_2$$. Without loss of generality assume that $$d(x,V_2) \leq d(y,V_1)$$. Then, $$d(x,y) \leq d(x,z) + d(z,y) < d(x,V_2)/3 + d(y,V_1)/3 \leq \frac 23 d(y,V_1),$$ which contradicts the definition of $$d(y,V_1)$$. Therefore $$U_1\cap U_2 = \emptyset$$.