[proofplan] We construct a sequence of unit vectors whose pairwise distances are uniformly bounded below. The construction uses [Riesz's lemma](/theorems/1222), proved here in the form needed, applied to the finite-dimensional spans of the previously chosen vectors. Since a compact [metric space](/page/Metric%20Space) is sequentially compact, compactness of the closed unit ball would force this sequence to have a convergent subsequence. The uniform separation prevents even a Cauchy subsequence, giving the contradiction. [/proofplan]

[step:Record the two compactness and separation facts needed for the construction]We first record two elementary facts. Throughout the proof, $\mathbb K$ denotes the scalar field of $X$, so $\mathbb K$ is either $\mathbb R$ or $\mathbb C$, and $\mathbb N=\{1,2,3,\dots\}$. [claim:Finite-dimensional subspaces of normed spaces are closed] Let $(X,\|\cdot\|_X)$ be a [normed vector space](/page/Normed%20Vector%20Space) and let $M\subset X$ be finite-dimensional. Then $M$ is closed in $X$. [/claim] [proof] We prove the claim by induction on $d:=\dim M$. If $d=0$, then $M=\{0\}$, and $\{0\}$ is closed because the norm map $x\mapsto \|x\|_X$ is continuous and $\{0\}=\{x\in X:\|x\|_X=0\}$. Assume the result holds for all subspaces of dimension at most $d-1$, and let $\dim M=d\ge 1$. Choose a subspace $N\subset M$ with $\dim N=d-1$ and choose $e\in M\setminus N$. By the induction hypothesis, $N$ is closed in $X$. Since $e\notin N$, define \begin{align*} \delta:=\operatorname{dist}(e,N):=\inf_{z\in N}\|e-z\|_X. \end{align*} Because $N$ is closed and $e\notin N$, we have $\delta>0$. Let $(y_k)_{k=1}^{\infty}$ be a sequence in $M$ converging in $X$ to some $y\in X$. For each $k\in\mathbb N$, write uniquely \begin{align*} y_k=z_k+a_k e \end{align*} with $z_k\in N$ and $a_k\in\mathbb R$ in the real case, or $a_k\in\mathbb C$ in the complex case. For $k,l\in\mathbb N$, the vector $z_k-z_l$ lies in $N$, so by the definition of $\delta$, \begin{align*} \|y_k-y_l\|_X=\|(z_k-z_l)+(a_k-a_l)e\|_X\ge |a_k-a_l|\delta. \end{align*} Since $(y_k)_{k=1}^{\infty}$ converges, it is Cauchy in $X$, and therefore $(a_k)_{k=1}^{\infty}$ is a [Cauchy sequence](/page/Cauchy%20Sequence) in the scalar field. Let $a$ denote its scalar limit. Then \begin{align*} z_k=y_k-a_k e\to y-ae \end{align*} in $X$. Since $N$ is closed, $y-ae\in N$. Hence $y=(y-ae)+ae\in M$. Thus every sequence in $M$ that converges in the metric induced by $\|\cdot\|_X$ has its limit in $M$. Because normed vector spaces are metric spaces, the sequential characterization of closed subsets in metric spaces gives that $M$ is closed. [/proof] [claim:Riesz separation lemma for a closed proper subspace] Let $(X,\|\cdot\|_X)$ be a normed [vector space](/page/Vector%20Space), let $M\subsetneq X$ be a closed linear subspace, and let $\alpha\in(0,1)$. Then there exists $u\in X$ such that $\|u\|_X=1$ and \begin{align*} \operatorname{dist}(u,M):=\inf_{m\in M}\|u-m\|_X>\alpha. \end{align*} [/claim] [proof] Choose $y\in X\setminus M$. Since $M$ is closed, define \begin{align*} d:=\operatorname{dist}(y,M)=\inf_{m\in M}\|y-m\|_X. \end{align*} Then $d>0$. Since $\alpha\in(0,1)$, the number $d/\alpha$ is strictly larger than $d$. By the definition of the infimum, choose $m_0\in M$ such that \begin{align*} d\le \|y-m_0\|_X<\frac{d}{\alpha}. \end{align*} Define \begin{align*} u:=\frac{y-m_0}{\|y-m_0\|_X}. \end{align*} Then $\|u\|_X=1$. For any $m\in M$, the vector $m_0+\|y-m_0\|_X m$ lies in $M$, since $M$ is a linear subspace. Hence \begin{align*} \|u-m\|_X=\frac{\|y-m_0-\|y-m_0\|_X m\|_X}{\|y-m_0\|_X}\ge \frac{d}{\|y-m_0\|_X}>\alpha. \end{align*} Taking the infimum over $m\in M$ gives $\operatorname{dist}(u,M)>\alpha$. [/proof][/step]

[guided]The proof needs two auxiliary facts, and both are included here to avoid hiding the key compactness mechanism behind references. First, [finite-dimensional subspaces are closed](/theorems/1225). The only point requiring care is that $X$ itself need not be complete. The proof proceeds by induction on the dimension. The zero-dimensional case is closed because $\{0\}$ is the zero set of the continuous norm map. For the induction step, write a $d$-dimensional subspace as \begin{align*} M=N+\operatorname{span}\{e\} \end{align*} where $\dim N=d-1$ and $e\notin N$. By induction, $N$ is closed. Define \begin{align*} \delta:=\operatorname{dist}(e,N)=\inf_{z\in N}\|e-z\|_X. \end{align*} Since $N$ is closed and $e\notin N$, the distance is positive. Now take a convergent sequence $(y_k)_{k=1}^{\infty}$ in $M$ and write \begin{align*} y_k=z_k+a_k e \end{align*} with $z_k\in N$ and scalar $a_k$. For two indices $k,l\in\mathbb N$, the difference has the form \begin{align*} y_k-y_l=(z_k-z_l)+(a_k-a_l)e. \end{align*} Because $z_k-z_l\in N$, the definition of $\delta$ gives \begin{align*} \|y_k-y_l\|_X\ge |a_k-a_l|\delta. \end{align*} Thus the scalar sequence $(a_k)_{k=1}^{\infty}$ is Cauchy whenever $(y_k)_{k=1}^{\infty}$ is Cauchy. Let $a$ be its scalar limit. Then \begin{align*} z_k=y_k-a_k e\to y-ae \end{align*} in $X$. Since $N$ is closed, $y-ae\in N$, and hence $y=(y-ae)+ae\in M$. This proves $M$ is closed. Second, Riesz's separation lemma supplies a unit vector away from a closed proper subspace. Let $M\subsetneq X$ be closed and choose $y\in X\setminus M$. Define \begin{align*} d:=\operatorname{dist}(y,M). \end{align*} Closedness gives $d>0$. For a fixed $\alpha\in(0,1)$, choose $m_0\in M$ with \begin{align*} d\le \|y-m_0\|_X<\frac{d}{\alpha}. \end{align*} Normalize the residual by setting \begin{align*} u:=\frac{y-m_0}{\|y-m_0\|_X}. \end{align*} Then $\|u\|_X=1$. For every $m\in M$, the vector $m_0+\|y-m_0\|_X m$ remains in $M$, so the definition of $d$ gives \begin{align*} \|u-m\|_X=\frac{\|y-m_0-\|y-m_0\|_X m\|_X}{\|y-m_0\|_X}\ge \frac{d}{\|y-m_0\|_X}>\alpha. \end{align*} Taking the infimum over all $m\in M$ yields $\operatorname{dist}(u,M)>\alpha$. This is exactly the separation estimate needed to build a sequence with no Cauchy subsequence.[/guided]

[step:Construct unit vectors separated from all previous choices] Choose $u_1\in X$ with $\|u_1\|_X=1$, which is possible because $X$ is infinite-dimensional and hence contains a nonzero vector. We construct a sequence $(u_n)_{n=1}^{\infty}$ in $X$ inductively. Suppose $u_1,\dots,u_n$ have been chosen with $\|u_j\|_X=1$ for every $j\in\{1,\dots,n\}$. Define the finite-dimensional subspace \begin{align*} M_n:=\operatorname{span}\{u_1,\dots,u_n\}\subset X. \end{align*} By the finite-dimensional closedness claim, $M_n$ is closed in $X$. Since $X$ is infinite-dimensional, $M_n\neq X$, so $M_n$ is a closed proper linear subspace. Apply the Riesz separation lemma with $\alpha=1/2$ to obtain $u_{n+1}\in X$ such that \begin{align*} \|u_{n+1}\|_X=1 \end{align*} and \begin{align*} \operatorname{dist}(u_{n+1},M_n)>\frac{1}{2}. \end{align*} This completes the induction and defines a sequence $(u_n)_{n=1}^{\infty}$ of unit vectors in $X$. [/step]

[step:Show the constructed sequence has no Cauchy subsequence] Let $m,n\in\mathbb N$ with $m<n$. Since $u_m\in M_{n-1}$, the separation property for $u_n$ gives \begin{align*} \|u_n-u_m\|_X\ge \operatorname{dist}(u_n,M_{n-1})>\frac{1}{2}. \end{align*} Thus every two distinct terms of the sequence are separated by more than $1/2$ whenever the larger index is used in the separation estimate. Let $(u_{n_j})_{j=1}^{\infty}$ be any subsequence, where $n_1<n_2<\dots$. For all $i<j$, \begin{align*} \|u_{n_j}-u_{n_i}\|_X>\frac{1}{2}. \end{align*} Therefore no subsequence of $(u_n)_{n=1}^{\infty}$ is Cauchy in the norm metric. In particular, no subsequence converges in the norm topology, since every convergent sequence in a metric space is Cauchy. [/step]

[step:Use metric compactness to force a convergent subsequence] Set \begin{align*} K:=\overline{B}_X(0,1)=\{u\in X:\|u\|_X\le 1\}. \end{align*} Each $u_n$ belongs to $K$ because $\|u_n\|_X=1$. Assume, for contradiction, that $K$ is compact in the norm topology. The restriction of the declared metric $d_X:X\times X\to\mathbb R$ to $K\times K$ is a metric inducing the subspace norm topology on $K$. We now use compactness of this metric space to extract a convergent subsequence from any sequence in $K$. Indeed, let $(x_j)_{j=1}^{\infty}$ be a sequence in $K$. If no point of $K$ were a cluster point of the sequence, then for every $x\in K$ there would exist $r_x>0$ such that the open ball \begin{align*} B_K(x,r_x):=\{y\in K:d_X(x,y)<r_x\} \end{align*} contains only finitely many terms of the sequence. The family $\{B_K(x,r_x):x\in K\}$ is an [open cover](/page/Open%20Cover) of $K$. Compactness gives points $x_1,\dots,x_N\in K$ such that \begin{align*} K\subset \bigcup_{i=1}^{N}B_K(x_i,r_{x_i}). \end{align*} The union on the right contains only finitely many terms of the sequence, contradicting that every term $x_j$ lies in $K$ and there are infinitely many indices $j\in\mathbb N$. Hence some $x\in K$ is a cluster point. Choosing inductively $j_q>j_{q-1}$ with \begin{align*} d_X(x_{j_q},x)<\frac{1}{q} \end{align*} gives a subsequence $(x_{j_q})_{q=1}^{\infty}$ converging to $x$ in $K$. Applying this compact metric [sequential compactness](/page/Sequential%20Compactness) argument to the sequence $(u_n)_{n=1}^{\infty}$ in $K$, we obtain a norm-convergent subsequence. This contradicts the previous step, which proved that $(u_n)_{n=1}^{\infty}$ has no norm-convergent subsequence. [/step]

[step:Conclude that the closed unit ball is not compact] The contradiction shows that the assumption of compactness of $K=\overline{B}_X(0,1)$ in the norm topology is false. Therefore the closed unit ball of an infinite-dimensional normed vector space is not compact in the norm topology. [/step]