[proofplan]
We embed the closed unit ball of $X^*$ isometrically into $\overline{B}_{\mathcal{L}(X,Y)}$ via rank-one operators $f \mapsto f(\cdot)y_0$, reducing the problem to the non-compactness of $\overline{B}_{X^*}$. Since $X$ is infinite-dimensional, so is $X^*$ (by the Hahn-Banach theorem), and [Riesz's Characterisation of Finite-Dimensional Normed Spaces](/theorems/1224) shows $\overline{B}_{X^*}$ is not norm-compact. An isometric copy of a non-compact set cannot live inside a compact set.
[/proofplan]
[step:Embed $\overline{B}_{X^*}$ isometrically into $\overline{B}_{\mathcal{L}(X,Y)}$ via rank-one operators]
Fix $y_0 \in Y$ with $\|y_0\|_Y = 1$ (which exists since $Y$ is non-trivial). Define the map
\begin{align*}
\Psi: X^* &\to \mathcal{L}(X, Y) \\
f &\mapsto f(\cdot) \, y_0,
\end{align*}
so that $\Psi(f)$ is the rank-one operator $x \mapsto f(x) y_0$. This map is linear and isometric:
\begin{align*}
\|\Psi(f)\|_{\mathcal{L}(X,Y)} = \sup_{\|x\|_X \le 1} \|f(x) y_0\|_Y = \sup_{\|x\|_X \le 1} |f(x)| \cdot \|y_0\|_Y = \|f\|_{X^*}.
\end{align*}
In particular, $\Psi$ maps $\overline{B}_{X^*} = \{f \in X^* : \|f\|_{X^*} \le 1\}$ isometrically onto the subset $\Psi(\overline{B}_{X^*}) \subset \overline{B}_{\mathcal{L}(X,Y)}$.
[guided]
The idea is to find a copy of an infinite-dimensional unit ball sitting inside $\overline{B}_{\mathcal{L}(X,Y)}$. The simplest approach is to restrict attention to rank-one operators: for a fixed non-zero $y_0 \in Y$, the operator $x \mapsto f(x) y_0$ is determined entirely by the functional $f \in X^*$. After normalising $\|y_0\| = 1$, the correspondence $f \mapsto f(\cdot) y_0$ becomes an isometry.
We verify this isometry. For $f \in X^*$:
\begin{align*}
\|\Psi(f)\|_{\mathcal{L}(X,Y)} = \sup_{\|x\|_X \le 1} \|f(x) y_0\|_Y = \sup_{\|x\|_X \le 1} |f(x)| \cdot \|y_0\|_Y = \|f\|_{X^*} \cdot 1 = \|f\|_{X^*}.
\end{align*}
Linearity of $\Psi$ is immediate: $\Psi(\alpha f + \beta g) = (\alpha f + \beta g)(\cdot) y_0 = \alpha f(\cdot) y_0 + \beta g(\cdot) y_0 = \alpha \Psi(f) + \beta \Psi(g)$.
Therefore $\Psi: X^* \to \mathcal{L}(X, Y)$ is a linear isometric embedding, and in particular $\Psi(\overline{B}_{X^*})$ is a norm-closed subset of $\overline{B}_{\mathcal{L}(X,Y)}$ (the image of a complete set under an isometry is complete, hence closed).
[/guided]
[/step]
[step:Show $X^*$ is infinite-dimensional and conclude non-compactness]
Since $X$ is infinite-dimensional, $X^*$ is also infinite-dimensional. To see this, suppose for contradiction that $\dim X^* = n < \infty$, so $X^* \cong \mathbb{R}^n$. The canonical embedding $J: X \to X^{**}$, defined by $J(x)(f) = f(x)$, is injective by the [Hahn-Banach Theorem](/theorems/879) (for any $x \ne 0$, there exists $f \in X^*$ with $f(x) \ne 0$, so $J(x) \ne 0$). But $X^{**} = (X^*)^* \cong (\mathbb{R}^n)^* \cong \mathbb{R}^n$, so $J$ is an injection from $X$ into $\mathbb{R}^n$, forcing $\dim X \le n < \infty$ — contradicting the hypothesis.
Since $X^*$ is infinite-dimensional, by [Riesz's Characterisation of Finite-Dimensional Normed Spaces](/theorems/1224), the closed unit ball $\overline{B}_{X^*}$ is not norm-compact. Since $\Psi: \overline{B}_{X^*} \to \Psi(\overline{B}_{X^*})$ is an isometry, $\Psi(\overline{B}_{X^*})$ is not norm-compact either (compactness is preserved and reflected by isometries). Since $\Psi(\overline{B}_{X^*})$ is a closed subset of $\overline{B}_{\mathcal{L}(X,Y)}$, and a closed subset of a compact space is compact, $\overline{B}_{\mathcal{L}(X,Y)}$ cannot be norm-compact.
[guided]
The argument has two parts: establishing $\dim X^* = \infty$, and transferring non-compactness through the isometry $\Psi$.
**$X^*$ is infinite-dimensional.** The [Hahn-Banach Theorem](/theorems/879) guarantees that $X^*$ separates points of $X$: for every non-zero $x \in X$, there exists $f \in X^*$ with $f(x) \ne 0$. This means the canonical embedding $J: X \to X^{**}$, $J(x)(f) = f(x)$, is injective. If $\dim X^* = n$, then $\dim X^{**} = \dim (X^*)^* = n$ (since $X^*$ is finite-dimensional), and $J: X \hookrightarrow X^{**} \cong \mathbb{R}^n$ would force $\dim X \le n$, contradicting $\dim X = \infty$.
**Non-compactness.** By [Riesz's Characterisation of Finite-Dimensional Normed Spaces](/theorems/1224), a normed space has compact closed unit ball if and only if it is finite-dimensional. Since $\dim X^* = \infty$, the ball $\overline{B}_{X^*}$ is not norm-compact.
Now $\Psi(\overline{B}_{X^*})$ is isometric to $\overline{B}_{X^*}$, and isometries preserve all metric properties — in particular, they preserve and reflect compactness (a subset $K$ of a metric space is compact if and only if every sequence in $K$ has a convergent subsequence with limit in $K$; isometries preserve convergence and limits). Therefore $\Psi(\overline{B}_{X^*})$ is not compact.
Finally, $\Psi(\overline{B}_{X^*})$ is a norm-closed subset of $\overline{B}_{\mathcal{L}(X,Y)}$ (since $\Psi$ is an isometric embedding of the Banach space $X^*$, its image is complete, hence closed). If $\overline{B}_{\mathcal{L}(X,Y)}$ were compact, every closed subset would be compact, and in particular $\Psi(\overline{B}_{X^*})$ would be compact — a contradiction.
[/guided]
[/step]
