(a) $C$ is non-empty and closed, $U$ is a component of $X-C$ . If $C=X$ then $U$ is empty otherwise it is not. So, suppose it is not. Theorem 24.1 tells us that $X$ and all rays and intervals in $X$ are connected. Then $x$ and $y$ are in the same component of $X-C$ iff $[x,y]$ does not intersect $C$ . If $C$ is bounded below by a point not in $C$ , then let $c$ be the greatest lower bound of $C$ . Since $C$ is closed, $c\in C$ (otherwise, let $x\in C$ , $c<x$ , i.e. $c$ is not the largest element of $X$ , and every basis neighborhood of $c$ has a from $(a,b)$ where $b>c$ , but then there is an element $x'\in (c,b)\cap C$ , i.e. $c$ cannot be an interior point of $X-C$ ). Hence, if there is a component containing a lower bound of $C$ then it is of the form $(-\infty,c)$ . Similarly, if there is a component containing an upper bound of $C$ then it is of the form $(c,+\infty)$ . Now, suppose that $U$ is bounded below and above by points in $C$ . Let $c$ be its greatest lower bound and $c'$ be its least upper bound. Neither $c$ nor $c'$ belongs to $U$ (essentially, for the same reason). But for every $c<x<y<c'$ there are points $c<a<x<y<b<c'$ such that $a,b\in U$ , therefore, $x,y\in[a,b]\subseteq U$ . Hence, $U=(c,c')$ .(b) Suppose $x$ is a limit point of $C$ . If $x$ is in a component then, using (a), the component is a neighborhood of $x$ that contains only one point in $C$ , therefore, since the space is Hausdorff, $x$ must be equal to that point. Suppose now, $x\in A$ . Then $x\notin X-A\cup B$ and each neighborhood of $x$ must contain infinite number of points in $C$ . Suppose $a<b<c$ are such points. $b$ lies within an interval with endpoints in $A$ and $B$ that contains neither $a$ nor $c$ . Therefore, $x$ is a limit point of $B$ , but $B$ is closed, so $x\in A\cap B$ . This contradicts the assumption that they are disjoint. Similarly, $x$ cannot lie within $B$ . To sum up, only points in $C$ can be limit points of $C$ , i.e. $C$ is closed.(c) Using (a) and (b) we conclude that the components of $X-C$ are open intervals or rays. Then $V$ is an interval that does not contain points in $C$ . Suppose $a\in A\cap V$ , $b\in B\cap V$ and $a<b$ . Let $b'$ be the greatest lower bound of points in $[a,b]\cap B$ . Then, as before, $b'\in B$ and $[a,b')$ contains no points in $B$ . Let $a'$ be the lowest upper bound of points in $[a,b']\cap A$ . Then $a'\in A$ , $a'<b'$ and $(a',b')$ has no points in $A\cup B$ . Therefore, it is a component of $X-A\cup B$ with the endpoints in both $A$ and $B$ , and must have a point in $C$ . Contradiction.Conclusion. For $A$ choose those components of $X-C$ that contains points in $A$ (they do not contain points in $B$ ), and for $B$ choose all other components. Since $C$ intersects neither $A$ nor $B$ , we have open disjoint sets separating $A$ and $B$ .