 Solution

Suppose otherwise and choose linearly independent eigenvectors $v_n$ with $T{v}_n={\lambda }_n{v}_n$ and $|{\lambda }_n|>a$. Put $F_n=\mathrm{span}(v_1,\dots ,v_n)$. Each $F_n$ is finite-dimensional and $T$-invariant. Moreover $(T-{\lambda }_{n}I)F_n\subseteq F_{n-1}$, including when some eigenvalues coincide.

Apply the supplied distance fact inside the Banach space $F_n$ to choose $x_n\in F_n$ with $\Vert x_n\Vert =1$ and $\mathrm{dist}(x_n,F_{n-1})=1$. For $m<n$, both $(T-{\lambda }_{n}I)x_n$ and $T{x}_m$ belong to $F_{n-1}$. ConsequentlyThus $(T{x}_n)$ has no Cauchy subsequence, although $(x_n)$ is bounded. This contradicts the compact operator property. Only finitely many linearly independent eigenvectors can have eigenvalue modulus greater than $a$.