Difficulty: 3

Contrapositively, suppose $\{V_n\}$ is a cover with no finite subcover. Then for $W_n = \bigcup_{m \le n} V_n$, each $X \setminus W_n$ is nonempty. So for each $n$, choose $x_n \in X \setminus W_n$. The sequence $\{x_n\}$ cannot have a converging subsequence: If $p \in X$, then $p \in V_m$ for some $m$ and so $x_n \notin V_m$ for all $n > m$.

Let $\{K_n\}$ be a countable compact cover of $X$ of which for all $x \in X$, some $K_n$ is a compact nbd of $x$. Let $K$ be compact. For each $x \in K$, let $n_x$ be the least element of which $K_n$ is a compact nbd. Then $\{\text{int}(K_{n_x})\}$ is a cover of $K$, which must have a finite subcover, and so finitely many $K_n$. Meaning $A \subseteq \bigcup_{n \le m} K_n$ for big enough $m$.

So define $K'_m = \bigcup_{n \le m} K_n$. Clearly $\{K'_m\}$ is a countable compact cover of $X$ and any compact is contained in some $K'_m$ by above.

We just have to argue a $k$-netork is a network: Let $\mathcal{N}$ be a $k$-network. Singletons are compact. So in a $k$-network, for each $x$ in an open set $U$, we can find $\mathcal{N}^*_x \subseteq \mathcal{N}$ with $\{x\} \subseteq \bigcup \mathcal{N}^*_x \subseteq U$ (we don’t need $\mathcal{N}^*_x$ to be finite here, so no need for the axiom of choice, just take every set of $\mathcal{N}$ containing $x$). Thus,

$$U = \bigcup_{x \in U} \{x\} \subseteq \bigcup_{x \in U} \mathcal{N}^*_x \subseteq U \ \implies \ \text{$U$ is a union of sets in $\mathcal{N}$, so it is a network}$$

Let $A$ be an infinite set. Contrapositively, suppose every limit point of $A$ is not an $\omega$-accumulation point. If $p$ were to be a limit point of $A$, there exists a nbd $V$ of $p$ of which $V \cap A = \{ a_1, \dots, a_n \}$ is finite. But now for each $a_k$, there’s $V_k$, a nbd of $p$, so that $a_k \notin V_k$ (by being $T_1$). Then $W = V \cap V_1 \cap \cdots \cap V_n$ is a nbd of $p$ with $W \cap A = \emptyset$. So $A$ is an infinite set with no limit points.

Let $f : X \to [0, 1]$ with $f(a) = 0$ and $f(b) = 1$. Take $V_a = f^{-1}([0, 1/3))$ and $V_b = f^{-1}((2/3, 1])$. Then $f(\overline{V_a}) \subseteq \overline{f(V_a)} \subseteq [0, 1/3]$ and $f(\overline{V_b}) \subseteq \overline{f(V_b)} = [2/3, 1]$. This proves $\overline{V_a} \cap \overline{V_b} = \emptyset$.

Let $\{N_n\}$ be the countable network and $A \subseteq X$. Ignore the network sets that don’t intersect $A$ and choose $x_n \in N_n \cap A$. Then $\{x_n\}$ has to be dense in $A$: For any $x \in A \cap V$, $x \in N_n \cap A \subseteq A \cap V$ for some $n$ so that $x_n \in A \cap V$.

If any path connected component is open, then there can only be one (as they form a partition of $X$, disproving being connected). Take the path connected component $P$ of $x$. For $y \in P$, let $U$ be a connected nbd. Then any $z \in U$ is path connected to $x$ (just concatenate the paths $x \sim y$ and $y \sim z$). So $U \subseteq P$.

Let $f : X \to Y \subseteq \R$ be a homeomorphism. $\R$ is $T_1$, so $Y \simeq X$ is as well. Define order on $X$ by $x < y$ iff $f(x) < f(y)$. If $U$ is a nbd of $x$, then $f(x) \in f(U)$ and $f(U) = V \cap Y$, where $I = (f(x) - \delta, f(x) + \delta) \subseteq V$ for some $\delta > 0$. Let $J = f^{-1}(I) \subseteq U$. Suppose $a,b \in J$ and $a < z < b$. Then $f(a) < f(z) < f(b)$ with $f(a),f(b) \in I$, so $f(z) \in I$ and $z \in J$ as we wished, to prove $J$ is order-convex.

Take any $x$ and a nbd $U$ which is $T_2$. We wish to find a neighborhood $V$ not containing some other $y$. If $y \notin U$, just take $V = U$. Otherwise, take $x \in V \subseteq U$ with $y \notin V$ since $U$ is $T_1$. But then $V$ is also open in $X$, so we’re done.

Let $x \sim y$ iff they are indistinguishable. The equivalent classes $[x]$ form a basis for a topology which must be finer than $X$ (if $U$ is a nbd of $x$, it must contain all $y \sim x$). It suffices to show each $[x]$ is an open set. But $Y = X \setminus [x]$ is compact, and any points of $Y$ are distinguishable from $x$, so it’s possible to find $x \in U$ and $Y \subseteq V$ nbds with $U \cap V = \emptyset$ (this is analogous to the result that for $T_2$ spaces, any point $x$ and a compact $K$ can be separated)