[proofplan] Write $A:=T(\overline{B}_H(0,1))$. Compactness of $T$ means that $A$ is relatively compact in $K$, that is, its closure in $K$ is compact. The forward implication follows by covering the compact set $\overline{A}$ with open $\varepsilon$-balls and extracting a finite subcover. Conversely, the assumed finite $\varepsilon$-ball covers say exactly that $A$ is totally bounded; since $K$ is complete, a standard diagonal subsequence argument shows that every sequence in $\overline{A}$ has a convergent subsequence, hence $\overline{A}$ is compact. [/proofplan] [step:Translate compactness of $T$ into relative compactness of the image of the unit ball] Define the subset $A\subset K$ by \begin{align*} A:=T(\overline{B}_H(0,1)). \end{align*} By the definition of a compact linear operator between normed spaces, $T$ is compact if and only if $A$ has compact closure in $K$. Thus the theorem is equivalent to the assertion that $\overline{A}^{\,K}$ is compact if and only if $A$ is contained in finitely many open balls of radius $\varepsilon$ for every $\varepsilon>0$. [/step] [step:Use compactness of the closure to obtain finite ball covers] Assume that $T$ is compact. Then $C:=\overline{A}^{\,K}$ is compact in $K$. Fix $\varepsilon>0$. The family of open balls \begin{align*} \{B_K(y,\varepsilon):y\in C\} \end{align*} is an open cover of $C$. Since $C$ is compact, there exist $N\in\mathbb N$ and points $y_1,\dots,y_N\in C$ such that \begin{align*} C\subset \bigcup_{j=1}^{N}B_K(y_j,\varepsilon). \end{align*} Because $A\subset C$ and $C\subset K$, the same points $y_1,\dots,y_N\in K$ satisfy \begin{align*} T(\overline{B}_H(0,1))=A\subset \bigcup_{j=1}^{N}B_K(y_j,\varepsilon). \end{align*} This proves the required finite covering property. [/step] [step:Build a Cauchy subsequence from the finite covers] Assume conversely that, for every $\varepsilon>0$, the set $A$ is contained in a finite union of open balls of radius $\varepsilon$ in $K$. We prove that every sequence in $A$ has a convergent subsequence in $K$. Let $(a_n)_{n=1}^{\infty}$ be a sequence in $A$. For each $m\in\mathbb N$, choose a finite set $F_m\subset K$ such that \begin{align*} A\subset \bigcup_{y\in F_m}B_K(y,1/m). \end{align*} Since $F_1$ is finite, one of the balls $B_K(y,1)$ with $y\in F_1$ contains infinitely many terms of $(a_n)$. Choose an infinite increasing index set $I_1\subset\mathbb N$ such that $(a_n)_{n\in I_1}$ lies in one such ball of radius $1$. Inductively, suppose $I_m\subset\mathbb N$ is infinite. Since $F_{m+1}$ is finite, one of the balls $B_K(y,1/(m+1))$ with $y\in F_{m+1}$ contains infinitely many terms $(a_n)_{n\in I_m}$. Choose an infinite subset $I_{m+1}\subset I_m$ such that all terms $(a_n)_{n\in I_{m+1}}$ lie in that ball. Now choose indices $n_m\in I_m$ inductively so that \begin{align*} n_1<n_2<\cdots<n_m<\cdots . \end{align*} For $q\ge p$, both indices $n_p$ and $n_q$ belong to $I_p$, because the sets satisfy \begin{align*} I_q\subset I_p. \end{align*} The terms $(a_n)_{n\in I_p}$ lie in a single open ball of radius $1/p$, so the triangle inequality gives \begin{align*} \|a_{n_q}-a_{n_p}\|_K<\frac{2}{p}. \end{align*} Hence $(a_{n_m})_{m=1}^{\infty}$ is a [Cauchy sequence](/page/Cauchy%20Sequence) in $K$. Since $K$ is a [Hilbert space](/page/Hilbert%20Space), it is complete, so there exists $a\in K$ such that \begin{align*} a_{n_m}\to a \end{align*} in the norm of $K$. [guided] The point of the finite covers is that they let us repeatedly trap infinitely many sequence terms in smaller and smaller balls. Let $(a_n)_{n=1}^{\infty}$ be a sequence in $A$. For each integer $m\ge 1$, the hypothesis with $\varepsilon=1/m$ gives a finite set $F_m\subset K$ satisfying \begin{align*} A\subset \bigcup_{y\in F_m}B_K(y,1/m). \end{align*} Start with the cover of radius $1$. Since there are only finitely many balls and the sequence has infinitely many terms, at least one ball in this cover contains infinitely many terms. Choose an infinite increasing set $I_1\subset\mathbb N$ such that all terms $(a_n)_{n\in I_1}$ lie in a single open ball of radius $1$. Now repeat the same argument inside the already chosen infinite subsequence. Suppose $I_m$ is an infinite set of indices. The radius $1/(m+1)$ cover consists of finitely many balls, and infinitely many terms $(a_n)_{n\in I_m}$ must be distributed among them. Therefore one of those balls contains infinitely many of these terms. Choose an infinite subset $I_{m+1}\subset I_m$ such that all terms $(a_n)_{n\in I_{m+1}}$ lie in that single open ball of radius $1/(m+1)$. This produces nested infinite sets \begin{align*} I_1\supset I_2\supset I_3\supset \cdots . \end{align*} Choose indices $n_m\in I_m$ inductively with \begin{align*} n_1<n_2<\cdots<n_m<\cdots . \end{align*} We claim that $(a_{n_m})_{m=1}^{\infty}$ is Cauchy. Fix $p\in\mathbb N$ and take $q\ge p$. Since the index sets are nested, $n_q\in I_q\subset I_p$, and also $n_p\in I_p$. By construction, all terms with indices in $I_p$ lie in one open ball of radius $1/p$. Therefore there is some center $c_p\in K$ such that \begin{align*} \|a_{n_p}-c_p\|_K<\frac{1}{p} \end{align*} and \begin{align*} \|a_{n_q}-c_p\|_K<\frac{1}{p}. \end{align*} Applying the triangle inequality in $K$ gives \begin{align*} \|a_{n_q}-a_{n_p}\|_K\le \|a_{n_q}-c_p\|_K+\|c_p-a_{n_p}\|_K<\frac{2}{p}. \end{align*} Given any $\delta>0$, choose $p$ so large that $2/p<\delta$. Then the displayed estimate holds for all $q\ge p$, so $(a_{n_m})_{m=1}^{\infty}$ is Cauchy. Because $K$ is a Hilbert space, it is complete as a normed space. Hence the Cauchy subsequence converges in $K$. [/guided] [/step] [step:Pass from sequences in $A$ to compactness of the closure] We now show that $C:=\overline{A}^{\,K}$ is compact. Let $(c_n)_{n=1}^{\infty}$ be a sequence in $C$. For each $n\in\mathbb N$, since $c_n\in\overline{A}^{\,K}$, there exists $a_n\in A$ such that \begin{align*} \|a_n-c_n\|_K<\frac{1}{n}. \end{align*} By the previous step, the sequence $(a_n)_{n=1}^{\infty}$ has a subsequence $(a_{n_m})_{m=1}^{\infty}$ converging to some $a\in K$. Then \begin{align*} \|c_{n_m}-a\|_K\le \|c_{n_m}-a_{n_m}\|_K+\|a_{n_m}-a\|_K. \end{align*} The first term tends to $0$ because $n_m\to\infty$, and the second term tends to $0$ by convergence of $(a_{n_m})$. Hence $c_{n_m}\to a$ in $K$. Because each $c_{n_m}\in C$ and $C$ is closed in $K$, the limit $a$ belongs to $C$. Thus every sequence in $C$ has a subsequence converging to a point of $C$. In metric spaces, [sequential compactness](/page/Sequential%20Compactness) is equivalent to compactness, so $C$ is compact. [/step] [step:Conclude the equivalence for the operator] Under the finite covering hypothesis, the preceding step proves that \begin{align*} \overline{T(\overline{B}_H(0,1))}^{\,K} \end{align*} is compact in $K$. Therefore $T(\overline{B}_H(0,1))$ is relatively compact in $K$. By the definition of compact linear operator, $T$ is compact. Together with the forward implication, this proves the equivalence. [/step]