= Solution
The <weighted binary-sequence metric> is finite because $0\leq d(\epsilon,\eta)\leq\sum_{n\geq1}2^{-n}=1$. It is symmetric and nonnegative. If $d(\epsilon,\eta)=0$, every nonnegative summand is zero, so every coordinate agrees; conversely agreement makes the sum zero. For three sequences, apply $|\epsilon_n-\zeta_n|\leq|\epsilon_n-\eta_n|+|\eta_n-\zeta_n|$ in each coordinate and sum. This proves the <triangle inequality>, so \b[$d$ is a <metric>].
If two sequences differ in one of their first $n$ coordinates, their distance is at least $2^{-n}$. Thus the <open ball> $B_d(\eta,2^{-n})$ is contained in $C_n(\eta)$. For every $\eta\in C_n(\epsilon)$, the latter cylinder equals $C_n(\eta)$, so each <cylinder set> is open in the <metric topology>. Every set in the specified topology, being a union of <cylinder sets>, is therefore metric-open.
Conversely, agreement in the first $n$ coordinates gives
$$
d(\epsilon,\eta)\leq\sum_{k>n}2^{-k}=2^{-n}.
$$
Given a metric-open set containing $\epsilon$, choose a ball $B_d(\epsilon,r)$ inside it, then choose $n$ with $2^{-n}<r$. The bound gives $C_n(\epsilon)\subseteq B_d(\epsilon,r)$, establishing the required prefix-neighbourhood condition. \b[The two topologies coincide.] The strict inequality $2^{-n}<r$ avoids incorrectly treating the closed tail bound as a strict ball bound.
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