 Solution

Start Brownian motions at arbitrary $x,y\in {\mathbb{R}}^d$. Use the successive meeting times from part b, but after coordinate $i$ meets, drive that coordinate of the second process with the first process's increments forever. The Strong Markov property shows that each marginal remains a $d$-dimensional Brownian motion. By part b every coordinate is eventually locked, so the resulting coordinatewise coalescing coupling of Brownian motions has an almost surely finite coalescence time $T$.

Because $f$ is bounded and harmonic, Dynkin formula for Brownian motion shows that $f(B_t^x)$ and $f(B_t^y)$ are bounded martingales. ThereforeSince $T<\infty$ almost surely, the right side tends to zero. Thus $f(x)=f(y)$ for all $x,y$, proving the Brownian coupling proof of the harmonic Liouville theorem.