A nonzero compact scalar identity forces finite dimension

The scalar operator $c^{-1}{I}_H$ satisfies $\mathrm{\parallel }{c}^{-1}{I}_{H}x\mathrm{\parallel }=\mathrm{\mid }c{\mathrm{\mid }}^{-1}\mathrm{\parallel }x\mathrm{\parallel }$ for every $x\in H$, so it is bounded with bound $\mathrm{\mid }c{\mathrm{\mid }}^{-1}$ so its operator norm is at most $\mathrm{\mid }c{\mathrm{\mid }}^{-1}$, including when $H=\{0\}$.

The closed unit ball $\bar{B}$ is closed in $H$: if $\mathrm{\parallel }x\mathrm{\parallel }>1$, put $r=\mathrm{\parallel }x\mathrm{\parallel }-1>0$. Whenever $\mathrm{\parallel }y-x\mathrm{\parallel }<r$, [F4] gives $\mathrm{\parallel }y\mathrm{\parallel }\ge \mathrm{\parallel }x\mathrm{\parallel }-\mathrm{\parallel }y-x\mathrm{\parallel }>1$, so the open ball of radius $r$ about $x$ is disjoint from $\bar{B}$. Its complement is therefore open.

The composite $(c{I}_H)\circ (c^{-1}{I}_H)$ is compact by [F2], applied with the compact operator $c{I}_H$ and the bounded operator $c^{-1}{I}_H$; this composite is the identity $I_H$.

By [F1] applied to $I_H$, the closure of $I_H(\bar{B})=\bar{B}$ in $H$ is compact. By step 1.2 the ball $\bar{B}$ is closed, so this closure is $\bar{B}$ itself; hence $\bar{B}$ is a compact subset of $H$.

By [F3] applied to the normed space $H$, compactness of $\bar{B}$ means that $H$ admits an ordered basis of finite length, that is, $H$ is finite dimensional.