A compact subspace of metric space with no isolated points. Suppose $X$ is a metric space which is connected and has no isolated points. Then I want to show that it will have some non-empty compact subspace $Y$ such that $Y$ has no isolated points. I don't know if my assertion is correct. If not, please give counterexample. Also, can we find non-empty subspace $Y$ such that $Y$ is both compact and connected. 
 A: Counterexample. Let $X$ be a Bernstein subset of $\mathbb R^2$, i.e., a subset of $\mathbb R^2$ such that both $X$ and $\mathbb R^2\setminus X$ meet every uncountable closed subset of $X$. (Such sets exist, assuming the axiom of choice.) Then $X$ is a connected metric space with no isolated points, and every nonempty compact subspace of $X$ has an isolated point.
Why is $X$ connected? Assume for a contradiction that $X=X_1\cup X_2$ where $X_1,X_2$ are nonempty and relatively closed in $X$, and $X_1\cap X_2=\emptyset$. Taking closures in $\mathbb R^2$ we have $\overline{X_1}\cup\overline{X_2}=\overline X=\mathbb R^2$, and $\overline{X_1}\cap\overline{X_2}$ is disjoint from $X$, whence $\overline{X_1}\cap\overline{X_2}$ is countable. Choose points $x_1\in X_1$ and $x_2\in X_2$. Let $\mathcal A$ be an uncountable collection of internally disjoint arcs from $x_1$ to $x_2$. Since $x_1,x_2\notin\overline{X_1}\cap\overline{X_2}$, some arc $A\in\mathcal A$ is disjoint from $\overline{X_1}\cap\overline{X_2}$. But then $A\cap\overline{X_1}$ and $A\cap\overline{X_2}$ are two disjoint nonempty subsets of $A$ whose union is $A$, which is impossible since $A$ is connected.
