Here is the theorem
As a reminder, a component is the maximal connected subset of an open set that contains a certain point $x$.
My problem is where he states that $B(x)$ is connected and concludes that it is in $T$.
Here's one line of reasoning:
We know there exists a ball $B(x)$ completely contained in $S$ but we can't guarantee that it contains only points in $T$ (it may contain points from other components). If it is completely contained in $T$ we're done, if not we consider the ball of radius $r/2$ centered around the same point and so on. If the process stops at some radius, we're done. If it doesn't, it means that a ball of arbitrarily small radius around $x$ contains points in other components which shows that the component $T$ contains only one point which is considered a closed set. How can we prove that the process stops?
Best Answer
The key result is: If $A,B$ are connected sets and $A \cap B$ is non empty then $A \cup B$ is connected.
If $T$ is a component it is connected. Suppose $x \in T$, then since $S$ is open, some open ball $B(x,\epsilon)$ is also contained in $S$. The open ball is connected (in fact path connected). Hence $T \cup B(x,\epsilon)$ is connected and since $T$ is maximal we have $T \cup B(x,\epsilon) \subset T$ and so $B(x,\epsilon) \subset T$ and so $T$ is open.