= Solution
Put $A=\{a,b\}$ and let $\tau_x^+=\min\{t\geq1:X_t=x\}$. The conductance-hitting identity for an <electrical network> is
$$
\mathbb P_x(\tau_D<\tau_x^+)
=\frac{1}{c(x)R_{\mathrm{eff}}(x,D)},
\qquad c(x)=\sum_yc(x,y),
$$
where the vertices of $D$ are wired together. It follows by taking the hitting probability of $D$ before returning to $x$ as a voltage and computing its total current out of $x$.
Split the walk into successive excursions from $x$. The first excursion that hits $A$ determines whether $a$ or $b$ is hit first. Its conditional probability of hitting $a$ is at most the probability that an arbitrary excursion hits $a$, divided by the probability that it hits $A$. Hence
$$
\begin{aligned}
\mathbb P_x(\tau_a<\tau_b)
&\leq
\frac{\mathbb P_x(\tau_a<\tau_x^+)}
{\mathbb P_x(\tau_A<\tau_x^+)}\\
&=\boxed{\frac{R_{\mathrm{eff}}(x,A)}{R_{\mathrm{eff}}(x,a)}}.
\end{aligned}
$$
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