= Solution
\b[False under the hypotheses as printed.] Nonzero coefficients do not ensure the <two-column dominant-subspace condition>.
For a counterexample, take $n=3$, eigenvalues
$$
(\lambda_1,\lambda_2,\lambda_3)=(1,2,-2),
$$
and an orthonormal eigenbasis $w_1,w_2,w_3$ for which
$$
e_1=\frac{-2w_1+w_2+w_3}{\sqrt6},
\qquad
e_2=\frac{w_1+w_2+w_3}{\sqrt3}.
$$
The 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, but
$$
b_2c_3-b_3c_2=0.
$$
Indeed, with $t=1/\sqrt2$,
$$
A^ke_1-tA^ke_2
$$
is a nonzero multiple of $w_1$, while the remaining direction in
$\operatorname{span}(A^ke_1,A^ke_2)$ is
$$
v_k=\frac{w_2+(-1)^kw_3}{\sqrt2}.
$$
Thus
$$
\operatorname{span}(A^ke_1,A^ke_2)
=\operatorname{span}(w_1,v_k).
$$
By 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 linear operator>[compression] of $A$ to this space. Since
$$
w_1^TAw_1=1,
\qquad
v_k^TAv_k=\frac{2+(-2)}2=0,
\qquad
w_1^TAv_k=0,
$$
we have
$$
\operatorname{Spec}(B_k)=\{0,1\}
$$
for every relevant $k$, rather than $\{-2,2\}$. Hence the requested <Hausdorff distance>[Hausdorff convergence] does not hold.
For completeness, the intended statement becomes true if one adds
$$
\Delta=b_{n-1}c_n-b_nc_{n-1}\ne0.
$$
Let $Z_k$ consist of the first two columns of $\overline Q_{k-1}$. The <Accumulated QR factorization identity> gives
$$
\operatorname{col}(Z_k)
=\operatorname{span}(A^ke_1,A^ke_2).
$$
After division by $|\lambda_n|^k$, every component along
$w_1,\ldots,w_{n-2}$ tends to zero because of the strict <spectral gap>, while $\Delta\ne0$ makes the two surviving dominant components independent. Therefore these two-dimensional subspaces converge to the <dominant invariant subspace>
$$
E=\operatorname{span}(w_{n-1},w_n).
$$
There are two-by-two orthogonal matrices $U_k$ such that
$$
Z_kU_k\longrightarrow W_*=(w_{n-1},w_n).
$$
Because
$$
B_k=Z_k^TAZ_k,
$$
up to the harmless index convention at $k=0$, we obtain
$$
U_k^TB_kU_k
\longrightarrow
W_*^TAW_*
=\operatorname{diag}(\lambda_{n-1},\lambda_n)
$$
in the <matrix 2-norm>. Orthogonal similarity preserves the <spectrum>, and the stated spectral perturbation bound now gives
$$
\boxed{d_H\!\left(\operatorname{Spec}(B_k),
\{\lambda_{n-1},\lambda_n\}\right)\longrightarrow0.}
$$
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