[proofplan] We refine the given sequence scale by scale using the finite $2^{-n}$-nets and the assumed infinite pigeonhole selection principle. At each scale we keep an infinite subsequence contained in one net ball, and we arrange the subsequences to be nested. A diagonal choice then produces a single subsequence whose sufficiently late terms lie in the same small ball at every prescribed scale. The triangle inequality makes this diagonal subsequence Cauchy, and completeness turns it into a convergent subsequence in $X$. [/proofplan] [step:Apply the pigeonhole selection at the first scale] Let \begin{align*} x:\mathbb{N} \to X \end{align*} be an arbitrary sequence. Apply the assumed selection principle with $n=1$ to the sequence $x$. There exist a center $c_1 \in E_1$ and a strictly increasing map \begin{align*} i_1:\mathbb{N} \to \mathbb{N} \end{align*} such that \begin{align*} x_{i_1(m)} \in B(c_1,2^{-1}) \end{align*} for every $m \in \mathbb{N}$. [/step] [step:Iteratively refine the surviving subsequence at each smaller scale] We construct, by induction on $k \in \mathbb{N}$, centers $c_k \in E_k$ and strictly increasing maps \begin{align*} i_k:\mathbb{N} \to \mathbb{N} \end{align*} such that the image of $i_{k+1}$ is contained in the image of $i_k$ and \begin{align*} x_{i_k(m)} \in B(c_k,2^{-k}) \end{align*} for every $m \in \mathbb{N}$. The case $k=1$ was constructed in the previous step. Suppose that $i_k$ has been constructed. Define the surviving sequence \begin{align*} z^{(k)}:\mathbb{N} \to X, \qquad m \mapsto x_{i_k(m)}. \end{align*} Apply the selection principle with $n=k+1$ to $z^{(k)}$. There exist $c_{k+1} \in E_{k+1}$ and a strictly increasing map \begin{align*} \phi_{k+1}:\mathbb{N} \to \mathbb{N} \end{align*} such that \begin{align*} z^{(k)}_{\phi_{k+1}(m)} \in B(c_{k+1},2^{-(k+1)}) \end{align*} for every $m \in \mathbb{N}$. Define \begin{align*} i_{k+1}:\mathbb{N} \to \mathbb{N}, \qquad m \mapsto i_k(\phi_{k+1}(m)). \end{align*} Because both $i_k$ and $\phi_{k+1}$ are strictly increasing, their composition $i_{k+1}$ is strictly increasing. Its image is contained in the image of $i_k$ by construction, and \begin{align*} x_{i_{k+1}(m)} = z^{(k)}_{\phi_{k+1}(m)} \in B(c_{k+1},2^{-(k+1)}) \end{align*} for every $m \in \mathbb{N}$. This completes the induction. [/step] [step:Choose a diagonal subsequence with strictly increasing original indices] We now choose integers $m_k \in \mathbb{N}$ recursively so that the original indices \begin{align*} j_k := i_k(m_k) \end{align*} are strictly increasing. Let $m_1 := 1$ and $j_1 := i_1(m_1)$. Suppose $m_k$ and $j_k$ have been chosen. Since $i_{k+1}:\mathbb{N} \to \mathbb{N}$ is strictly increasing, its values are unbounded in $\mathbb{N}$. Hence there exists $m_{k+1} \in \mathbb{N}$ such that \begin{align*} i_{k+1}(m_{k+1}) > j_k. \end{align*} Set \begin{align*} j_{k+1} := i_{k+1}(m_{k+1}). \end{align*} Thus $j_1 < j_2 < \cdots$. Define the diagonal subsequence \begin{align*} y:\mathbb{N} \to X, \qquad k \mapsto x_{j_k}. \end{align*} Because the map $k \mapsto j_k$ is strictly increasing, $y$ is a subsequence of $x$. [/step] [step:Show that every tail of the diagonal subsequence lies in one small net ball] Fix $r \in \mathbb{N}$. If $k \ge r$, then the image of $i_k$ is contained in the image of $i_r$, because the images are nested at every refinement stage. Hence there exists an integer $a_{r,k} \in \mathbb{N}$ such that \begin{align*} j_k = i_k(m_k) = i_r(a_{r,k}). \end{align*} By the defining property of the scale-$r$ subsequence, \begin{align*} y_k = x_{j_k} = x_{i_r(a_{r,k})} \in B(c_r,2^{-r}). \end{align*} Therefore every term $y_k$ with $k \ge r$ lies in the same ball $B(c_r,2^{-r})$. [guided] Fix a scale $r \in \mathbb{N}$. The point of the nested construction is that once a term survives to a later scale, it has also survived every earlier scale. Formally, the induction gave \begin{align*} \operatorname{im}(i_{k+1}) \subset \operatorname{im}(i_k) \end{align*} for every $k \in \mathbb{N}$. Therefore, if $k \ge r$, repeated containment gives \begin{align*} \operatorname{im}(i_k) \subset \operatorname{im}(i_r). \end{align*} The diagonal index $j_k$ was defined by $j_k := i_k(m_k)$, so $j_k \in \operatorname{im}(i_k)$. Since $\operatorname{im}(i_k) \subset \operatorname{im}(i_r)$, there exists $a_{r,k} \in \mathbb{N}$ such that \begin{align*} j_k = i_r(a_{r,k}). \end{align*} The scale-$r$ construction says that every point with index in the image of $i_r$ lies in the selected ball $B(c_r,2^{-r})$. Thus \begin{align*} y_k = x_{j_k} = x_{i_r(a_{r,k})} \in B(c_r,2^{-r}) \end{align*} for every $k \ge r$. This is the essential diagonal observation: the $k$th diagonal term was chosen from the $k$th refined subsequence, but the nesting forces it to remain inside all earlier chosen balls. Hence, after the $r$th term, the entire tail of the diagonal subsequence lies in the single ball $B(c_r,2^{-r})$. [/guided] [/step] [step:Use the triangle inequality to prove the diagonal subsequence is Cauchy] Let $\varepsilon > 0$ be given. Since $2^{1-r} \to 0$ as $r \to \infty$, choose $r \in \mathbb{N}$ such that \begin{align*} 2^{1-r} < \varepsilon. \end{align*} If $k,\ell \ge r$, then the previous step gives \begin{align*} y_k \in B(c_r,2^{-r}) \quad \text{and} \quad y_\ell \in B(c_r,2^{-r}). \end{align*} By the triangle inequality in the [metric space](/page/Metric%20Space) $(X,d)$, \begin{align*} d(y_k,y_\ell) \le d(y_k,c_r) + d(c_r,y_\ell). \end{align*} Because both terms lie in the open ball $B(c_r,2^{-r})$, we have \begin{align*} d(y_k,c_r) < 2^{-r} \quad \text{and} \quad d(c_r,y_\ell) < 2^{-r}. \end{align*} Therefore \begin{align*} d(y_k,y_\ell) < 2^{-r} + 2^{-r} = 2^{1-r} < \varepsilon. \end{align*} Thus $y$ is a [Cauchy sequence](/page/Cauchy%20Sequence) in $X$. [/step] [step:Apply completeness to obtain a convergent subsequence in $X$] The sequence \begin{align*} y:\mathbb{N} \to X \end{align*} is a Cauchy subsequence of the original sequence $x$. Since $(X,d)$ is complete, every Cauchy sequence in $X$ converges to a point of $X$. Hence there exists $y_\infty \in X$ such that \begin{align*} d(y_k,y_\infty) \to 0 \end{align*} as $k \to \infty$. Therefore the original sequence $x$ has a subsequence converging to a limit in $X$. [/step]