Let $(X,d)$ be a metric space, show that if $A \subset X$ is connected, $B \subset X$ with $A \subset B \subset cl(A)$ then B is connected.
Let's assume that B is not connnected, then by definition there exist two open, disjunct subsets $U, V \subset X$ such that $ B \subset U \cup V, B \cap V \neq \emptyset, B \cap U \neq \emptyset$. Let $b_1 \in B \cap U, b_2 \in B \cap V$. Then since $B \subset cl(A) \Rightarrow \forall \epsilon \gt 0: K(b_1, \epsilon) \cap A \neq \emptyset$ and $\forall \epsilon \gt 0: K(b_2, \epsilon) \cap A \neq \emptyset$
Now I'm not sure whether the next step is correct, to me it seems as if since $K(b_1, \epsilon) \cap A \neq \emptyset$ is true for all $\epsilon \gt 0$, there has to be an element $a_1 \in A : a_1 = b_1$ and the same logic would apply to b_2.
Then there would exist $U, V \subset X : A \subset U \cup V, A \cap V \neq \emptyset, A \cap U \neq \emptyset$, therefore A is not connected.
Is that correct?