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Here we will be using the notation $a \times b$ to denote the ordered pair having $a$ as its first coordinate and $b$ as its second coordinate.

Here is Example 2, Sec. 25, in the book Topology by James R. Munkres, 2nd edition:

The "topologist's since curve" $\overline{S}$ of the preceding section is a space that has a single component (since it is connected) and two path components. One path component is the curve $S$ and the other is the vertical interval $V = 0 \times [-1, 1]$. Note that $S$ is open in $\overline{S}$ but not closed, while $V$ is closed but not open.

If one forms a space from $\overline{S}$ by deleting all points of $V$ having rational second coordinates, one obtains a space that has only one component but uncountably many path components.

I'm unable to figure out the two claims made in the second paragraph.

How is it that, despite the removal of the set $0 \times \big( \mathbb{Q} \cap [-1, 1] \big)$ from $\overline{S}$, we are still left with a space that is connected? Apparently, this process of removing points from the vertical interval would leave holes in it!

As for the second claim, is it not true that the uncountably many path components will include those of the form $\{ 0 \times x \}$, where $x \in [-1, 1] \setminus \mathbb{Q}$? In short my question is, is the following subspace of $\mathbb{R}^2$ path connected? $$ V^\prime = 0 \times \left( [-1, 1] \cap \mathbb{Q}^c \right). $$

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For the first question, any point in $V$ is the limit of points in $S$. So the new space is contained between $S$ and $\overline S$ so is connected, as $S$ is connected.

For your second question, any non-trivial path in $0\times[-1,1]$ will contain a point with rational coordinates, so removing all these leaves a space where all path components have just one point.

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