= Solution
Suppose otherwise and choose linearly independent <eigenvectors> $v_n$ with $Tv_n=\lambda_nv_n$ and $|\lambda_n|>a$. Put $F_n=\operatorname{span}(v_1,\ldots,v_n)$. Each $F_n$ is finite-dimensional and $T$-invariant. Moreover $(T-\lambda_nI)F_n\subseteq F_{n-1}$, including when some <eigenvalues> coincide.
Apply the supplied distance fact inside the <Banach space> $F_n$ to choose $x_n\in F_n$ with $\|x_n\|=1$ and $\operatorname{dist}(x_n,F_{n-1})=1$. For $m<n$, both $(T-\lambda_nI)x_n$ and $Tx_m$ belong to $F_{n-1}$. Consequently
$$
\|Tx_n-Tx_m\|\ge \operatorname{dist}(\lambda_nx_n,F_{n-1})=|\lambda_n|>a.
$$
Thus $(Tx_n)$ has no Cauchy <subsequence>, although $(x_n)$ is bounded. This contradicts the <compact operator> property. \b[Only finitely many linearly independent <eigenvectors> can have <eigenvalue> modulus greater than $a$.]
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