Every sequence in $X$ has a cluster point in $X$ $\implies$ Every open cover of $X$ has a finite subcover 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.
 A: Here is my attempt on proving the equivalence of these three statements. It would be great if someone helps me verify the other two attempts. Thank you so much @Paul Sinclair for your very kindness!


Theorem: 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 $X$.

My Attempt:

*

*Every open cover of $X$ has a finite subcover $\implies$ $X$ is totally bounded and complete

For $\epsilon > 0$, $\{\mathbb B(x,\epsilon) \mid x \in X\}$ is an open cover of $X$. Then there is a finite set $I \subseteq X$ such that $\cup_{x \in I} \mathbb B(x,\epsilon)$ covers $X$. Hence $X$ is totally bounded.
Let $(x_n)$ be a Cauchy sequence in $X$. Assume the contrary that $(x_n)$ does not converge to any point in $X$. Then $(x_n)$ has no cluster point in $X$. Thus, for any $x \in X$, there is a neighborhood $\mathcal U_x$ of $x$ such that $\mathcal U_x$ contains at most finitely many terms of $(x_n)$. Because $\{\mathcal U_x \mid x \in X\}$ is an open cover of $X$, there is a finite set $I \subseteq X$ such that $\bigcup_{x \in I} \mathcal U_x$ covers $X$. On the other hand, $\cup_{x \in I} \mathcal U_x$ contains at most finitely many terms of $(x_n)$. This contradiction shows that $(x_n)$ converges to some point in $X$, so $X$ is complete.


*$X$ is totally bounded and complete $\implies$ Every sequence in $X$ has a cluster point in $X$
Let $(x_n)$ be a sequence in $X$. We define the sequences $(y_k),(I_k)$ and a mapping $\varphi:\mathbb N_+ \to \mathbb N$ recursively as follows:
The case $k=1$:

*

*Because $X$ is totally bounded, $X$ is covered by finitely many balls $B_i=\mathbb B(y_1^i,1)$ where $y_1^i \in X$ for all $i= \overline{1,n_1}$.


*Then there exists $i$ such that infinitely many terms of $(x_n)$ belongs to $B_i$. If not, for all $i= \overline{1,n_1}$, $B_i$ contains at most finitely many terms of $(x_n)$, and so does $\bigcup_{i=1}^{n_1} B_i$. This contradicts the fact that $\bigcup_{i=1}^{n_1} B_i$ covers $X$.


*Let $y_{1} = y^{n_0}_{1}$ where $n_0$ is the least $i$ such that infinitely many terms of $(x_n)$ belongs to $B_i$. Let $I_1 = \{n \in \mathbb N \mid x_n \in \mathbb B(y_1,1)\}$ and $\varphi (1) = \min \{n \in \mathbb N \mid x_n \in \mathbb B(y_1,1)\}$.
The case $k=2$:

*

*Because $X$ is totally bounded, $X$ is covered by finitely many balls $B_i=\mathbb B(y_2^i,1/2)$ where $y_2^i \in X$ for all $i= \overline{1,n_2}$.


*Then there exists $i$ such that infinitely many terms of $(x_n)$ belongs to $B_i \cap \mathbb B(y_1,1)$. If not, for all $i= \overline{1,n_2}$, $B_i \bigcap  \mathbb B(y_1,1)$ contains at most finitely many terms of $(x_n)$, and so does $\bigcup_{i=1}^{n_1}\left [B_i \bigcap  \mathbb B(y_1,1)\right] = \left(\bigcup_{i=1}^{n_1} B_i\right) \bigcap \mathbb B(y_1,1) = \mathbb B(y_1,1)$. This contradicts the construction of $y_1$.


*Let $y_2 = y^{n_0}_2$ where $n_0$ is the least $i$ such that infinitely many terms of $(x_n)$ belongs to $B_i \cap \mathbb B(y_1,1)$. Let $I_2 = \{n \in \mathbb N \mid x_n \in \mathbb B(y_1,1) \bigcap \mathbb B(y_2,1/2)\}$ and $\varphi (2) = \min (I_{2} \setminus \{n \in \mathbb N \mid 1 \le n \le \varphi (1)\})$.
The inductive case:

*

*We have $X$ is covered by finitely many balls $B_i :=\mathbb B(y_{k+1}^i, 1/(k+1))$ where $y_{k+1}^i \in X$ for all $i= \overline{1,n_{k+1}}$.


*There exists $i$ such that $B_i \bigcap \left (\bigcap_{n=1}^k \mathbb B(y_n, 1/n) \right)$ contains infinitely many terms of $(x_n)$. If not, $\mathbb B_i \bigcap \left (\bigcap_{n=1}^k \mathbb B(y_n, 1/n) \right)$ contains at most finitely many terms of $(x_n)$ for all $i= \overline{1,n_{k+1}}$, and so does $\bigcup_{i=1}^{n_{k+1}}  \left [B_i \bigcap \left (\bigcap_{n=1}^k \mathbb B(y_n, 1/n) \right) \right] =  \left(\bigcup_{i=1}^{n_{k+1}} B_i \right) \bigcap \left (\bigcap_{n=1}^k \mathbb B(y_n, 1/n) \right) =$ $Y \bigcap \left (\bigcap_{n=1}^k \mathbb B(y_n, 1/n) \right)= \bigcap_{n=1}^k \mathbb B(y_n, 1/n)$. This contradicts the definition of $y_k$.


*Let $y_{k+1} = y^{n_0}_{k+1}$ where $n_0$ is the least $i$ such that infinitely many terms of $(x_n)$ belongs to $B_i \bigcap \left (\bigcap_{n=1}^k \mathbb B(y_n, 1/n) \right)$. Let $I_{k+1} = \{m \in \mathbb N \mid x_m \in \bigcap_{n=1}^{k+1} \mathbb B(y_n, 1/n)\}$ and $\varphi (k+1) = \min (I_{k+1} \setminus \{n \in \mathbb N  \mid 1 \le n \le \varphi (k)\})$.
By construction, we have

*

*$\varphi$ is strictly increasing and thus $(x_{\varphi (n)})$ is a subsequence of $(x_n)$.


*$x_{\varphi(n)} \in \mathbb B(y_k, 1/k)$ for all $k = \overline{1,n}$.
Next we prove that $(x_{\varphi (n)})$ is a Cauchy sequence. Given $\epsilon >0$, there is $N \in \mathbb N$ such that $1/N < \epsilon/2$. For all $n \ge N$, we have $x_{\varphi(n)} \in \mathbb B(y_N, 1/N)$ and $x_{\varphi (N)} \in \mathbb B(y_N, 1/N)$. Thus $\| x_{\varphi (n)} - y_N\| < 1/N$ and $\| y_N - x_{\varphi (N)}\| <1/N$. As such, $\| x_{\varphi (n)} - x_{\varphi (N)}\| \le \| x_{\varphi (n)} - y_N\| + \| y_N - x_{\varphi (N)}\|<$ $1/N +1/N < \epsilon$ for all $n > N$.
Because $X$ is complete, $(x_{\varphi (n)})$ converges to some $\bar x \in Y$. Hence $\bar x$ is a cluster point of $(x_n)$.


*Every sequence in $X$ has a cluster point in $X$ $\implies$ Every open cover of $X$ has a finite subcover

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 $i \in I$ such that $\mathbb B(a,r) \subseteq O_i$ if such $i$ exists, otherwise $f(a,r) =i_0$ for some fixed $i_0 \in I$. Let $J=f[A \times \mathbb Q_+]$. 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 $\{O_i \mid i \in I \}$ has a countable subcover $\{O_k \mid k \in \mathbb N\}$. 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 $n \in \mathbb N$. As such, $X_n:=\bigcap_{k=0}^n O^c_{k} = \left (\bigcup_{k=0}^n O_{k} \right)^c \neq \emptyset$ for all $n \in \mathbb N$. Moreover, $X_{n+1} \subseteq X_n$ and $X_n$ is closed in $X$ for all $n$. On the other hand, $\{O_k \mid k \in \mathbb N\}$ is a subcover of $X$, so $\bigcup_{k=0}^\infty O_k =X$ or equivalently $\bigcap_{n=0}^\infty X_n = \bigcap_{k=0}^\infty O^c_{k} = \emptyset$.


*By Axiom of Countable Choice, we define  the sequence $(x_n)$ in $X$ recursively by $x_n \in X_n$ for all $n \in \mathbb N$. 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 from 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 $\bar x \in \bigcap_{n=0}^\infty X_n = \emptyset$. This is a contradiction. Hence $\{O_i \mid i \in I \}$ has a finite subcover.
A: Assuming (iii) try to prove any open cover has a countable sub-cover first, i.e. some listable cover B_1,B_2,B_3,... . Then you can argue inductively that this can be reduced to a sub-cover B_1',B_2',B_3',... such that no B_i' is covered by the other sets. At this point you can pick x_1,x_2,x_3,... from each set but outside all the others. Either this sub-cover is finite or you have just constructed a sequence which can only accumulate to its limit x. However, x would have to be in one of the B_i'. This cannot be since only one of the x_j's is in this set. 
