Good evening, I'm trying to prove that
Let $X$ be a metric space. If every sequence in $X$ has a cluster point in $X$ $\implies$ every open cover of $X$ has a finite subcover.
Actually, I'm trying to prove
Let $X$ be a metric space. Prove that the following statements are equivalent.
(i) Every open cover of $X$ has a finite subcover.
(ii) $X$ is totally bounded and complete.
(iii) Every sequence in $X$ has a cluster point in $X$.
I've just proved (i) implies (ii) and (ii) implies (iii). The only remaining is (iii) implies (i).
Could you please verify whether my attempt is fine or contains logical gaps/errors? Any suggestion is greatly appreciated!
My attempt:
Let $d$ be the metric on $X$.
First, we prove that $X$ is totally bounded. Suppose the contrary that $X$ is not totally bounded.
Then there exists $r>0$ such that $X \not \subseteq \bigcup_{k=0}^{m} \mathbb{B}\left(x_{k}, r\right)$ for any finite set $\{x_{0}, \ldots, x_{m}\} \subseteq K .$ In particular, there exists $x_{0} \in K$ such that $X \not \subseteq \mathbb{B} (x_{0}, r) .$ Thus there exists $x_{1} \in \left(\mathbb{B}(x_{0}, r)\right)^c$. Since $X \not \subseteq \left (\mathbb{B}(x_{0}, r) \bigcup \mathbb{B}(x_{1}, r)\right)$, there exists $x_{2} \in \left(\mathbb{B}(x_{0}, r) \bigcup \mathbb{B}(x_{1}, r)\right)^c$. Continuing in this way and with Axiom of Countable Choice, there is a sequence $(x_n)$ in $X$ such that $x_{n+1} \in \left(\bigcup_{k=0}^n \mathbb{B}(x_{k}, r)\right)^c$. It follows that $x_{n+1} \notin \mathbb{B}(x_{k}, r)$ and thus $d(x_k, x_{n+1}) \ge r$ for all $k \le n$.
By our hypothesis, $(x_n)$ has a cluster point, i.e. there exists a subsequence $(x_{\psi(n)})$ of $(x_n)$ such that $x_{\psi(n)} \to \bar x$ as $n \to \infty$. It follows from $x_{\psi(n)} \to \bar x$ that there is $N \in \mathbb N$ such that $d(x_{\psi(n)},\bar x) < r/2$ for all $n \ge N$. Hence $d(x_{\psi(N)},\bar x) < r/2$ and $d(x_{\psi(N+1)},\bar x) < r/2$. It follows that $d(x_{\psi(N)}, x_{\psi(N+1)}) \le d(x_{\psi(N)},\bar x) + d(x_{\psi(N+1)},\bar x) < r/2 +r/2 = r$. This contradicts our construction of $(x_n)$. Hence $X$ is totally bounded.
Next, we prove that every open cover $\{O_i \mid i \in I \}$ of $X$ has a countable subcover.
Because $X$ is totally bounded, for each $n \ge 1$ there are finitely many $x^i_n$ such that $X = \bigcup_{i=0}^{k_n} \mathbb B (x_n^i , 1/n)$. Let $A = \bigcup_{n=1}^\infty \{x_n^0, \ldots, x_n^{k_n}\}$. Because $A$ is countable union of countable sets, $A$ is countable.
We define a mapping $f:A \times \mathbb Q_+ \to I$ by corresponding (with help from Axiom of Choice) $(a,r) \in A \times \mathbb Q_+$ with an $i$ such that $\mathbb B(a,r) \subseteq O_i$ if such $i$ exists, otherwise $f(a,r) =O_{i_0}$ for some $i_0 \in I$. Let $J=f[I]$. Because $A,\mathbb Q_+$ are countable, $A \times \mathbb Q_+$ is countable and so is $J$.
For $x\in X$, there exists some $j \in I$ such that $x \in O_j$. Because $O_j$ is open, there is $r>0$ such that $\mathbb B(x,r) \subseteq O_j$. Then we choose some $a \in A$ such that $d(a,x) < r/2$ and some $r' \in \mathbb Q$ such that $d(a,x) <r' < r/2$. It follows that $x \in \mathbb B(a,r') \subseteq \mathbb B(x,r) \subseteq O_j$ by triangle inequality. By the construction of $f$, $x \in O_{f(a,r')}$. As such, $\{O_i \mid i \in J\}$ is a countable subcover of $X$.
Finally, we prove that $X$ is compact. Let $\{O_i \mid i \in I \}$ be an open cover of $X$.
We've just proved that there exists a countable subcover $\{O_k \mid k \in \mathbb N\}$ of $\{O_i \mid i \in I \}$. Assume the contrary that $\{O_i \mid i \in I \}$ has no finite subcover. Then $X \not \subseteq \bigcup_{k =0}^n O_k$ for any $k \in \mathbb N$. As such, $X_n:=\bigcap_{k=0}^n O^c_{k} = \left (\bigcup_{k=0}^n O_{k} \right)^c \neq \emptyset$ for all $n \in \mathbb N$. On the other hand, $(O_k)_{k \in \mathbb N}$ is a subcover of $X$, so $\bigcup_{k=0}^\infty O_k =X$ or equivalently $\bigcap_{k=0}^\infty X_k = \bigcap_{k=0}^\infty O^c_{k} = \emptyset$.
By Axiom of Countable Choice, we define the sequence $(x_n)$ in $X$ recursively by $x_{n+1} \in X_n:= \bigcap_{k=0}^{n+1} O^c_{k}$ for all $n \in \mathbb N$. It follows that $X_{n+1} \subseteq X_n$ and that $X_n$ is closed in $X$ for all $n$. By hypothesis, $(x_n)$ has a cluster point $\bar x \in X$, i.e. there is a subsequence $(x_{\phi(n)})$ of $(x_n)$ such that $x_{\phi(n)} \to \bar x \in X$. It follows our construction of $(x_n)$ that $x_{\phi(n)} \in X_N$ for all $n \ge N$. Moreover, $X_N$ is closed, so $\bar x \in X_N$. Because this is true for all $N$, we have $\bar x \in \bigcap_{k=0}^\infty X_k = \emptyset$. This is a contradiction. Hence $\{O_i \mid i \in I \}$ has a finite subcover.