Probs. 3 (a & c), Sec. 27, in Munkres' TOPOLOGY, 2nd ed: The space $\mathbb{R}_K$ is not path connected and $[0, 1]$ is not compact in $\mathbb{R}_K$. Here is Prob. 3, Sec. 27, in the book Topology by James R. Munkres, 2nd edition:

Recall that $\mathbb{R}_K$ denotes $\mathbb{R}$ in the $K$-topology.
(a) Show that $[0, 1]$ is not compact as a subspace of $\mathbb{R}_K$.
(b) Show that $\mathbb{R}_K$ is connected. [Hint: $(-\infty, 0)$ and $(0, \infty)$ inherit their usual topologies as subspaces of $\mathbb{R}_K$.]
(c) Show that $\mathbb{R}_K$ is not path connected.

Here is my Math SE post on Part (b) of this particular problem.
My Attempt:

We recall that the set $K$ is given by
  $$ K = \left\{ \ 1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \ldots \ \right\}, \tag{Definition 0}$$
  and the $K$-topology on the set $\mathbb{R}$ of real numbers is the topology having as a basis all the open intervals $(a, b)$, where $a \in \mathbb{R}$, $b \in \mathbb{R}$, and $a < b$, together with all the sets of the form $(a, b) \setminus K$. And, set $\mathbb{R}$ with the $K$-topology is denoted by $\mathbb{R}_K$. [Please refer to Sec. 13 in Munkres, specifically the Definition preceding Lemma 13.4.]

Part (a):

In this part, we will be using Lemma 26.1 in Munkres.
Let $\mathscr{A}$ be the following covering of $[0, 1]$ by sets open in $\mathbb{R}_K$.
  $$ \mathscr{A} \colon= \big\{ \ (-1, 2) \setminus K \ \big\} \bigcup \left\{ \ \left( \frac{1}{n+1}, 2\right) \ \colon \ n \in \mathbb{N} \ \right\}. $$
  [Refer to (Definition 0) above.]
We note that there is no finite sub-collection of $\mathscr{A}$ that could possibly cover the entire closed interval $[0, 1]$.

PS: 
In order to verify the assertion in the preceding paragraph, let us consider any finite sub-collection of the open covering $\mathscr{A}$, say, the sub-collection
$$ \mathscr{A}^\prime \colon=  \big\{ \ (-1, 2) \setminus K \ \big\} \bigcup \left\{ \ \left( \frac{1}{n_1 + 1}, 2\right), \ldots, \left( \frac{1}{n_k + 1}, 2\right) \ \right\},  $$
where $n_1, \ldots, n_k$ are distinct natural numbers. Then we find that
$$ \bigcup_{S \in \mathscr{A}^\prime} S = \big( \, (-1, 2) \setminus K \, \big) \cup \left( \frac{1}{n_0+1}, 2 \right), $$
where 
$$ n_0 \colon= \max \left\{ n_1, \ldots, n_k \right\}. $$
Thus, for example,
$$ \frac{1}{n_0+2} \in [0, 1] \setminus \bigcup_{S \in \mathscr{A}^\prime} S. $$
Therefore no finite sub-collection $\mathscr{A}^\prime$ of $\mathscr{A}$ can cover $[0, 1]$ as a subspace of $\mathbb{R}_K$.

Hence $[0, 1]$ is not compact as a subspace of $\mathbb{R}_K$.

Is this proof correct? 
Part (c):
First, here is the definition of a topological space to be path connected.

Given points $x$ and $y$ of a topological space $X$, a path in $X$ from $x$ to $y$ is a continuous map $f \colon [a, b] \to X$ of some closed interval in the real line into $X$, such that $f(a) = x$ and $f(b) = y$. A topological space $X$ is said to be path connected if every pair of points of $X$ can be joined by a path in $X$.

Please refer to Sec. 24 in Munkres, specifically the Definition preceding Example 3.

Suppose that $\mathbb{R}_K$ is path connected. Then there is a path in $\mathbb{R}_K$ from  point $x = 0$ to point $y = 1$; that is, for some closed interval $[a, b]$ on the real line, there exists a continuous mapping $f \colon [a, b] \to \mathbb{R}_K$ such that $f(a) = 0$ and $f(b) = 1$. 
We note that the closed interval $[a, b]$ is both connected and compact as a subspace of the topological space $\mathbb{R}$ (i.e. the set $\mathbb{R}$ of real numbers with the standard, or usual, topology). [Please refer to Corollary 24.2 and Corollary 27.2 in Munkres.]
Now as the closed interval $[a, b]$ is both connected and compact as a subspace of $\mathbb{R}$ and as $f$ is a continuous mapping of $[a, b]$ into $\mathbb{R}_K$, so the image $f\big([a, b]\big)$ is also connected and compact as a subspace of $\mathbb{R}_K$. [Please refer to Theorems 23.5 and 26.5 in Munkres.]
Thus as a subspace of $\mathbb{R}_K$, the set $f\big( [a, b] \big)$ is both compact and connected, and $f(a) = 0$ and $f(b) = 1$. 
Let $y$ be any real number in the open interval $(0, 1)$. Now as $[a, b]$ is connected, as $f \colon [a, b] \to \mathbb{R}_K$ is continuous, and as $f(a) = 0$ and $f(b) = 1$, so there exists a real number $c \in (a, b)$ such that $f(c) = y$, by the intermediate-value theorem (i.e. Theorem 24.3 in Munkres). Thus we have shown that 
  $$ [0, 1] \subset f\big( [a, b] \big). \tag{1} $$

PS: The argument in the preceding paragraph is flawed, as has been pointed out in one of the comments below. So here is the correct argument:

Let $y$ be any real number in the open interval $(0, 1)$. Suppose that there is no $r \in [a, b]$ for which $f(r) = y$. 
As $(-\infty, y) \cup (y, \infty) = \mathbb{R}\setminus \{ y \}$ and as $y \not\in f\big([a, b]\big)$, so we find that
  $$ \big(\, (-\infty, y) \cap f\big([a, b]\big) \, \big) \cup \big( \, (y, +\infty)\cap f\big([a, b]\big) \, \big) = \big( \mathbb{R} \setminus \{y \} \big) \cap f\big([a, b]\big) = f\big([a, b]\big).   $$
Now as the unbounded open intervals $(-\infty, y)$ and $(y, +\infty)$ are open in $\mathbb{R}_K$, so the sets $(-\infty, y)\cap f\big([a, b]\big)$ and $(y, +\infty)\cap f\big([a, b]\big)$ are open in the subspace $f\big([a, b]\big)$ of $\mathbb{R}_K$. 
Moreover, the sets $(-\infty, y)\cap f\big([a, b]\big)$ and $(y, +\infty)\cap f\big([a, b]\big)$ are disjoint.
Thus the disjoint open sets $(-\infty, y)\cap f\big([a, b]\big)$ and $(y, +\infty)\cap f\big([a, b]\big)$ form a separation of $f\big([a, b]\big)$. But this contradicts the fact that $f\big([a, b]\big)$ is connected. 
Thus there does not exist any real number $y \in (0, 1)$ for which there is no $r \in [a, b]$ such that $f(r) = y$. That is, for any real number $y \in (0, 1)$, we have $y = f(r)$ for at least one $r \in [a, b]$. Moreover, as $f(a) = 0$ and $f(b) = 1$, so we conclude that
  $$ [0, 1] \subset f\big([a, b]\big). $$
Now as the closed interval $[0, 1]$ is a closed subset of $\mathbb{R}_K$ and as $[0, 1] \subset f\big( [a, b] \big)$, so 
  $$[0, 1] = [0, 1] \cap f \big( [a, b] \big),$$ and thus $[0, 1]$ is also a closed subset of $f \big( [a, b] \big)$. [Please refer to Theorem 17.2 in Munkres.]
And, as $f\big([a, b]\big)$ is compact and as $[0, 1]$ is a closed subspace of $f\big([a, b]\big)$, so $[0, 1]$ is also compact as a subspace of $f\big([a, b]\big)$, by Theorem 26.2 in Munkres. 
Finally, as $[0, 1]$ is a compact subspace of $f\big([a, b]\big)$ and as $f \big([a, b]\big)$ is a subspace of $\mathbb{R}_K$, so $[0, 1]$ is also compact as a subspace of $\mathbb{R}_K$. 
But this contradicts the fact $[0, 1]$ is NOT compact as a subspace of $\mathbb{R}_K$, as we have shown in Part (a) above.
Thus our supposition that $\mathbb{R}_K$ is path connected is wrong. Hence $\mathbb{R}_K$ is not path connected.

Is this proof correct? 
Are both proofs given above correct and clear enough in each and every detail? Or, are there any point(s) in either of these where there are issues or errors?
 A: It's easier to note that $K \subseteq [0,1]$ is closed and discrete and then $[0,1]$ is not compact, as $K$ is not. 
But the cover argument also works, but needs to be expanded a bit by an argument why there is no finite subcover of that given cover. 
The appeal to 24.3 (IVT) is not correct as we don't have the order topology in the image. But you can adapt its proof and show $[0,1] \subseteq f[[0,1]]$ that way. And then the same contradiction can be achieved by the non-compactness of $[0,1]\subseteq \Bbb R_K$.
As to (b), the hint is very relevant: $(-\infty ,0)$ has the same topology in $\mathbb{R}_K$ as it has in the normal reals (we only modify the topology around $0$) and so is connected and hence so is its closure (in $\mathbb{R}_K$ too this closed segment is the closure), and the same holds for $(0,\infty)$ and its closure $[0,\infty)$ and the union of these intersecting connected closed segments is $\mathbb{R}_K$ which is thus connected by standard theorems. 
Another way to see the last remarks on $\mathbb{R}_K$'s topology: we can see that topology slightly differently by looking at the closed sets: we add one new closed set $K= \{\frac1n: n \in \Bbb N\}$ to the closed sets of $\Bbb R$ in its usual topology and generate a new collection of closed sets (just as we do for open bases, we have closed bases too). As $K \cap (-\infty,0) = \emptyset$ we add no new closed sets in that subspace and as $K \cap (0,\infty) = (K \cup \{0\}) \cap (0, \infty)$, where the latter set is already closed in $(0,\infty)$ (as $K \cup \{0\}$ is Euclidean closed, even compact), we also add no new closed subsets to $(0,\infty)$ as well. Hence the fact that these open segments still have the Eculidean subspace topology and are thus connected (as "continua" in Munkres terms). Every basic neighbourhood of $0$, be it of the usual $(-r,r)$ form or of the $(-r,r)-K$ form, intersects both open segments so that the closures are as I said.
