[proofplan] We choose a maximal subset $N \subset S^{d-1}$ whose distinct points are separated by distance greater than $\varepsilon$. Maximality forces $N$ to be an $\varepsilon$-net, because any point not within distance $\varepsilon$ of $N$ could be added. The balls of radius $\varepsilon/2$ around points of $N$ are disjoint and lie inside the Euclidean ball of radius $1+\varepsilon/2$, so comparison of Lebesgue volumes gives the desired cardinality bound. [/proofplan]

[step:Choose a maximal $\varepsilon$-separated subset of the sphere]Call a subset $A \subset S^{d-1}$ strictly $\varepsilon$-separated if for every pair of distinct points $u,v \in A$, \begin{align*} |u-v| > \varepsilon. \end{align*} By [Zorn's lemma](/theorems/1226) applied to the partially ordered family of strictly $\varepsilon$-separated subsets of $S^{d-1}$ ordered by inclusion, there exists a maximal strictly $\varepsilon$-separated set $N \subset S^{d-1}$. Indeed, the union of any chain of strictly $\varepsilon$-separated subsets is again strictly $\varepsilon$-separated, because any two distinct points in the union lie together in one member of the chain.[/step]

[guided]We first need a separated set that cannot be enlarged. Define a subset $A \subset S^{d-1}$ to be strictly $\varepsilon$-separated if every two distinct points $u,v \in A$ satisfy \begin{align*} |u-v| > \varepsilon. \end{align*} The strict inequality is useful because it will imply disjointness of the open balls of radius $\varepsilon/2$ around the selected points. We order the collection of all strictly $\varepsilon$-separated subsets of $S^{d-1}$ by inclusion. To use Zorn's lemma, take any chain $\mathcal{C}$ in this collection and define \begin{align*} A_{\mathcal{C}} := \bigcup_{A \in \mathcal{C}} A. \end{align*} We check that $A_{\mathcal{C}}$ is strictly $\varepsilon$-separated. If $u,v \in A_{\mathcal{C}}$ are distinct, then there exist $A_1,A_2 \in \mathcal{C}$ with $u \in A_1$ and $v \in A_2$. Since $\mathcal{C}$ is a chain, either $A_1 \subset A_2$ or $A_2 \subset A_1$. Thus $u$ and $v$ both lie in one strictly $\varepsilon$-separated set, so \begin{align*} |u-v| > \varepsilon. \end{align*} Therefore every chain has an upper bound. Zorn's lemma gives a maximal strictly $\varepsilon$-separated subset $N \subset S^{d-1}$.[/guided]

[step:Use maximality to prove that $N$ is an $\varepsilon$-net] We claim that $N$ is an $\varepsilon$-net for $S^{d-1}$. Suppose, to the contrary, that there exists $x \in S^{d-1}$ such that $|x-u| > \varepsilon$ for every $u \in N$. Then $N \cup \{x\}$ is strictly $\varepsilon$-separated, contradicting the maximality of $N$. Hence for every $x \in S^{d-1}$ there exists $u \in N$ with $|x-u| \leq \varepsilon$. [/step]

[step:Compare the volumes of disjoint balls around the net points] For each $u \in N$, define the open Euclidean ball \begin{align*} B\left(u,\frac{\varepsilon}{2}\right) := \left\{y \in \mathbb{R}^d : |y-u| < \frac{\varepsilon}{2}\right\}. \end{align*} If $u,v \in N$ are distinct, then \begin{align*} B\left(u,\frac{\varepsilon}{2}\right) \cap B\left(v,\frac{\varepsilon}{2}\right) = \varnothing. \end{align*} Indeed, if $y$ belonged to the intersection, the triangle inequality would give \begin{align*} |u-v| \leq |u-y| + |y-v| < \frac{\varepsilon}{2}+\frac{\varepsilon}{2} = \varepsilon, \end{align*} contradicting strict $\varepsilon$-separation. Moreover, since $|u|=1$ for every $u \in N$, each point $y \in B(u,\varepsilon/2)$ satisfies \begin{align*} |y| \leq |y-u|+|u| < \frac{\varepsilon}{2}+1. \end{align*} Thus \begin{align*} \bigcup_{u \in N} B\left(u,\frac{\varepsilon}{2}\right) \subset B\left(0,1+\frac{\varepsilon}{2}\right). \end{align*} Let $\mathcal{L}^d$ denote the $d$-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) on $\mathbb{R}^d$, and define \begin{align*} \alpha_d := \mathcal{L}^d(B(0,1)). \end{align*} Thus $\alpha_d$ is the $d$-dimensional Lebesgue measure of the open unit ball in $\mathbb{R}^d$. By translation and scaling invariance of $\mathcal{L}^d$, \begin{align*} \mathcal{L}^d\left(B\left(u,\frac{\varepsilon}{2}\right)\right) = \alpha_d \left(\frac{\varepsilon}{2}\right)^d \end{align*} for every $u \in N$, and \begin{align*} \mathcal{L}^d\left(B\left(0,1+\frac{\varepsilon}{2}\right)\right) = \alpha_d \left(1+\frac{\varepsilon}{2}\right)^d. \end{align*} Since the balls $B(u,\varepsilon/2)$ are pairwise disjoint and contained in $B(0,1+\varepsilon/2)$, finite additivity for pairwise disjoint measurable sets gives, for every finite subset $F \subset N$, \begin{align*} |F|\,\alpha_d\left(\frac{\varepsilon}{2}\right)^d = \mathcal{L}^d\left(\bigcup_{u \in F} B\left(u,\frac{\varepsilon}{2}\right)\right) \leq \alpha_d\left(1+\frac{\varepsilon}{2}\right)^d. \end{align*} Because $\alpha_d>0$, this implies \begin{align*} |F| \leq \left(\frac{1+\varepsilon/2}{\varepsilon/2}\right)^d = \left(1+\frac{2}{\varepsilon}\right)^d. \end{align*} The right-hand side is finite, so $N$ cannot contain arbitrarily large finite subsets. Therefore $N$ is finite, and taking $F=N$ yields \begin{align*} |N| \leq \left(1+\frac{2}{\varepsilon}\right)^d. \end{align*} [/step]

[step:Conclude the existence of the required finite net] The set $N \subset S^{d-1}$ constructed above is finite, satisfies the covering property \begin{align*} \forall x \in S^{d-1}\ \exists u \in N \quad |x-u| \leq \varepsilon, \end{align*} and obeys the cardinality estimate \begin{align*} |N| \leq \left(1+\frac{2}{\varepsilon}\right)^d. \end{align*} Thus $N$ is an $\varepsilon$-net of $S^{d-1}$ with the asserted size bound. [/step]