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False under the hypotheses as printed. Nonzero coefficients do not ensure the two-column dominant-subspace condition.

For a counterexample, take $n=3$, eigenvaluesand an orthonormal eigenbasis $w_1,w_2,w_3$ for whichThe two displayed coefficient vectors are orthonormal and can be completed to an orthogonal matrix, so such a real symmetric matrix $A$ exists. Every $b_i$ and $c_i$ is nonzero, butIndeed, with $t=1/\sqrt{2}$,is a nonzero multiple of $w_1$, while the remaining direction in
$\mathrm{span}(A^k{e}_1,A^k{e}_2)$ isThusBy the simultaneous iteration interpretation of the QR algorithm, the leading two columns of the accumulated $Q$ factor span this same space. The leading block $B_k$ is therefore an orthogonal-coordinate representation of the compression of $A$ to this space. Sincewe havefor every relevant $k$, rather than $\{-2,2\}$. Hence the requested Hausdorff convergence does not hold.

For completeness, the intended statement becomes true if one addsLet $Z_k$ consist of the first two columns of ${\overline{Q}}_{k-1}$. The Accumulated QR factorization identity givesAfter division by $|{\lambda }_n{|}^k$, every component along
$w_1,\dots ,w_{n-2}$ tends to zero because of the strict spectral gap, while $\Delta \ne 0$ makes the two surviving dominant components independent. Therefore these two-dimensional subspaces converge to the dominant invariant subspaceThere are two-by-two orthogonal matrices $U_k$ such thatBecauseup to the harmless index convention at $k=0$, we obtainin the matrix 2-norm. Orthogonal similarity preserves the spectrum, and the stated spectral perturbation bound now gives