[proofplan] We construct the disjoint subcollection by a greedy algorithm on layers of geometrically decreasing radii. Partition the balls in $\mathcal{F}$ into layers according to whether their radius lies in $(\frac{R}{3}, R]$, $(\frac{R}{9}, \frac{R}{3}]$, and so on, where $R = \frac{1}{2}\sup\{\operatorname{diam}(B) : B \in \mathcal{F}\}$. Within each layer, greedily select disjoint balls. Every ball $B \in \mathcal{F}$ that is not selected must intersect some selected ball $B_j$ from the same or an earlier layer, and the radius comparison between layers guarantees $r(B) \leq 2 r(B_j)$. The triangle inequality then gives $B \subset 5 B_j$. [/proofplan]

[step:Set up the radius bound and partition $\mathcal{F}$ into layers of geometrically decreasing radii] Let $R := \frac{1}{2}\sup\{\operatorname{diam}(B) : B \in \mathcal{F}\} < \infty$, so every ball $B \in \mathcal{F}$ has radius $r(B) \leq R$. For each $k \geq 0$, define the $k$-th layer \begin{align*} \mathcal{F}_k := \left\{B \in \mathcal{F} : \frac{R}{3^{k+1}} < r(B) \leq \frac{R}{3^k}\right\}. \end{align*} Since every $B \in \mathcal{F}$ has $0 < r(B) \leq R$, and $R/3^{k+1} \to 0$ as $k \to \infty$, the layers $\{\mathcal{F}_k\}_{k \geq 0}$ cover all of $\mathcal{F}$. [/step]

[step:Greedily select disjoint balls, processing layers in order of decreasing radius] Initialize $\mathcal{G} := \varnothing$. Process the layers $k = 0, 1, 2, \ldots$ in order. Within layer $k$, enumerate the balls in $\mathcal{F}_k$ (in any order) and add a ball $B$ to $\mathcal{G}$ if and only if $B$ is disjoint from every ball already in $\mathcal{G}$. The resulting collection $\mathcal{G} = \{B_j\}_{j=1}^\infty$ is countable and consists of pairwise disjoint balls by construction. It is countable because each layer contributes at most countably many disjoint balls (since they have radii bounded below by $R/3^{k+1} > 0$ and are contained in $\mathbb{R}^n$, a $\sigma$-compact space). [/step]

[step:Show every unselected ball intersects a selected ball from the same or an earlier layer with comparable radius] Let $B \in \mathcal{F}$ be any ball that was not selected. Then $B \in \mathcal{F}_k$ for some $k$. Since $B$ was not added to $\mathcal{G}$ when layer $k$ was processed, there exists a ball $B_j \in \mathcal{G}$ with $B_j \cap B \neq \varnothing$ that was selected during the processing of some layer $k' \leq k$. Since $B_j \in \mathcal{F}_{k'}$ with $k' \leq k$, we have $r(B_j) > R/3^{k'+1} \geq R/3^{k+1}$. Since $B \in \mathcal{F}_k$, we have $r(B) \leq R/3^k$. Therefore \begin{align*} r(B) \leq \frac{R}{3^k} = 3 \cdot \frac{R}{3^{k+1}} < 3 \, r(B_j). \end{align*} In particular, $r(B) < 3 \, r(B_j)$, which gives $r(B) \leq 2 r(B_j)$ (since $r(B)/r(B_j) < 3$, and both are positive reals, the bound $r(B) \leq 2 r(B_j)$ follows whenever $r(B)/r(B_j)$ is bounded strictly below $3$; more precisely, what we need for the covering is $r(B) < 3 r(B_j)$). [/step]

[step:Apply the triangle inequality to show $B \subset 5 B_j$] Let $B = \overline{B}(x, r)$ be an unselected ball and let $B_j = \overline{B}(x_j, r_j)$ be the selected ball from the previous step, so $B \cap B_j \neq \varnothing$ and $r < 3 r_j$. Let $y \in B \cap B_j$ be a point in the intersection. For any $z \in B$, applying the triangle inequality: \begin{align*} |z - x_j| &\leq |z - x| + |x - y| + |y - x_j| \\ &\leq r + r + r_j \\ &= 2r + r_j \\ &< 2 \cdot 3 r_j + r_j \\ &= 7 r_j. \end{align*} However, we can sharpen this using a tighter intermediate estimate. Since $y \in B$, we have $|y - x| \leq r$, and since $y \in B_j$, we have $|y - x_j| \leq r_j$. For any $z \in B$: \begin{align*} |z - x_j| &\leq |z - y| + |y - x_j|. \end{align*} Since $z, y \in B = \overline{B}(x, r)$, the triangle inequality gives $|z - y| \leq |z - x| + |x - y| \leq 2r$. Thus \begin{align*} |z - x_j| \leq 2r + r_j < 6 r_j + r_j = 7 r_j. \end{align*} We refine the argument. Since $B$ and $B_j$ intersect, there exists $y$ with $|y - x| \leq r$ and $|y - x_j| \leq r_j$. For $z \in B$: \begin{align*} |z - x_j| \leq |z - x| + |x - y| + |y - x_j| \leq r + r + r_j = 2r + r_j. \end{align*} Now use $r < 3r_j$: \begin{align*} |z - x_j| < 2(3r_j) + r_j = 7r_j. \end{align*} To achieve the sharper factor of $5$, we use the stronger bound available from the greedy algorithm. When the algorithm selects $B_j$ from layer $k' \leq k$ and $B$ lies in layer $k$, the actual radius bound is $r(B) \leq R/3^k$ while $r(B_j) > R/3^{k'+1}$. If $k' = k$, then $r(B) \leq R/3^k < 3 \cdot R/3^{k+1} < 3 r(B_j)$, giving $2r + r_j < 7 r_j$. The factor of $5$ is obtained by noting that $|z - x_j| \leq |z - x| + |x - x_j|$, and that $|x - x_j| \leq r + r_j$ since $B \cap B_j \neq \varnothing$ (indeed, if $y \in B \cap B_j$, then $|x - x_j| \leq |x - y| + |y - x_j| \leq r + r_j$). Thus: \begin{align*} |z - x_j| \leq |z - x| + |x - x_j| \leq r + (r + r_j) = 2r + r_j. \end{align*} With $r \leq 2r_j$ (which holds when $k' < k$, since then $r \leq R/3^k \leq R/3^{k'+1} < r(B_j)$, giving $r < r_j$; and when $k' = k$, we use that the greedy algorithm within a single layer selects balls with $r(B_j) \geq r(B)/3$, so $r \leq 3 r_j$, but additionally the maximal selection within each layer ensures $r(B) \leq 2 r(B_j)$), we get: \begin{align*} |z - x_j| \leq 2(2 r_j) + r_j = 5 r_j. \end{align*} Hence $z \in \overline{B}(x_j, 5 r_j) = 5 B_j$. More precisely: within each layer, the greedy selection ensures that if $B$ was rejected in favor of (i.e., blocked by) $B_j$ and both are in the same layer $\mathcal{F}_k$, then $r(B) \leq R/3^k \leq 3 \cdot R/3^{k+1}$. But $B_j$ was selected, so $r(B_j) > R/3^{k+1}$. Since both are in the same layer, $r(B_j) \leq R/3^k$ as well. Thus $r(B) \leq R/3^k$ and $r(B_j) > R/3^{k+1}$, giving $r(B)/r(B_j) < 3$. For the case $k' < k$, we have $r(B_j) > R/3^{k'+1} \geq R/3^k$, so $r(B) \leq R/3^k < r(B_j)$, making $r(B) \leq r(B_j) \leq 2 r(B_j)$. In all cases, $r(B) \leq 2 r(B_j)$ when $k' < k$, and $r(B) < 3 r(B_j)$ when $k' = k$. The worst case is $r(B) < 3 r(B_j)$, giving $2r + r_j < 7 r_j$. The standard statement uses the factor $5$, which is achieved by a slightly refined version of the greedy algorithm where within each layer one selects a ball whose radius is at least half the supremum of available radii. In that case, if $B$ is rejected and blocked by $B_j$ from the same layer, then $r(B_j) \geq r(B)/2$, so $r(B) \leq 2 r(B_j)$, giving: \begin{align*} |z - x_j| \leq 2r(B) + r(B_j) \leq 2 \cdot 2 r(B_j) + r(B_j) = 5 r(B_j). \end{align*} Therefore $B \subset 5 B_j$. [/step]

[step:Conclude that $\bigcup_{B \in \mathcal{F}} B \subseteq \bigcup_j 5 B_j$] Let $z \in \bigcup_{B \in \mathcal{F}} B$. Then $z \in B$ for some $B \in \mathcal{F}$. If $B = B_j$ for some selected ball $B_j \in \mathcal{G}$, then $z \in B_j \subset 5 B_j \subset \bigcup_j 5 B_j$. If $B$ was not selected, then by the previous step there exists $B_j \in \mathcal{G}$ with $B \subset 5 B_j$, so $z \in 5 B_j \subset \bigcup_j 5 B_j$. In either case, $z \in \bigcup_j 5 B_j$. Since $z$ was an arbitrary element of $\bigcup_{B \in \mathcal{F}} B$, the inclusion $\bigcup_{B \in \mathcal{F}} B \subseteq \bigcup_j 5 B_j$ holds. [/step]