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.