= Solution
Assume $n\ge3$, so the corner entry is separate from the superdiagonal. For $A(\delta)$, the row radii are
$$
r_1=\delta+\epsilon,\qquad r_i=q=\tfrac14+\epsilon\ (2\le i<n),\qquad r_n=\tfrac14.
$$
Thus every <eigenvalue> lies in the union $\bigcup_iD(i,r_i)$. The discs with centers $2,\ldots,n$ remain mutually disjoint. The expanding disc centered at one meets the disc centered at $j$ exactly when $j-1\le\delta+\epsilon+r_j$.
Let $I_\delta=\{1\}\cup\{j\ge2:j-1\le\delta+\epsilon+r_j\}$. Then the union of discs indexed by $I_\delta$ contains exactly $|I_\delta|$ <eigenvalues>; every other disc contains exactly one. Every isolated <eigenvalue> is real by conjugation and obeys $|\lambda-j|\le r_j$. Once discs merge, the second <Gershgorin circle theorem> gives a bound and count for their whole component, rather than one <eigenvalue> per original disc. <Eigenvalue> branches coming from that component stay in its union, but may no longer be distinguished by individual discs. These statements describe the bounds through all merger thresholds as $\delta$ increases.
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