Connectivity of a subset of the topologist's sine curve I have a question about Example 2 of Section 25 (p.160) of Munkres's Topology.
Let $S$ be the following subset of the plane $\mathbb{R}^2$: $$S = \{ \ x \times \sin \tfrac{1}{x} \ \colon \ 0 < x \leq 1 \ \}.$$ 
Then $S$ is connected, being a continuous image of the connected subset $(0, 1]$ of $\mathbb{R}$, and the topologist's sine curve is the closure $\overline{S}$ in $\mathbb{R}^2$ of $S$ and is therefore also connected. 
Let $V = \{0 \} \times [-1,1]$. Then $\overline{S} = V \cup S$. 
Now Munkres states that the space obtained from $\overline{S}$ by deleting from $V$ all points $0 \times q$ such that $q \in \mathbb{Q}$, the set of rational numbers, is also connected.
How do we show that this is so? I know that the subspace $\mathbb{R} - \mathbb{Q}$ is not connected. I would prefer a rigorous argument, although an intuitive one would also be welcome.
 A: This is a special case of the following statement:
Suppose $A \subset X$ is connected. Then every subset $B \subset X$ such that $A \subset B \subset \bar{A}$, is also connected.
Proof: Suppose $U,V$ are open sets which form a separation of $B$. Since $B \subset \bar{A}$, we know that $U \cap A$ and $V \cap A$ are both nonempty. Therefore $U \cap A$, $V \cap A$ form a separation of $A$, contradicting the fact that $A$ is connected.
Now simply apply this statement when $A=S$ and $B = S \cup (\{0\} \times ([0,1] \backslash \mathbb{Q}))$.
A: Intuitively, even though $\{0\} \times ([0,1] \cap \Bbb R \setminus \Bbb Q)$ is not connected (it is in fact totally disconnected), when you attach the topologist's sine curve it ''glues'' these points together, just in a different manner than adding in the rationals would glue them together.
More formally, let $U$ be a closed and open subset of $\overline{S} \setminus (\{0\} \times ([0,1] \cap \Bbb Q))$.  Note that $U \cap S$ is closed and open in $S$, so it is either all of $S$ or disjoint from $S$.  In the former case, the fact that $U$ is closed means it also contains every point of $\{0\} \times ([0,1] \cap \Bbb R \setminus \Bbb Q$), and in the latter case the fact that it is open means it can contain no points of $\{0\} \times ([0,1] \cap \Bbb R \setminus \Bbb Q)$.
