= Solution
Fix $x\in(-1,1)$ and start at $(x,y)$ with $y>1$. Write the independent Brownian coordinates as $x+W_t^1$ and $y+W_t^2$. Let
$$
H=\inf\{t:|x+W_t^1|=1\},
\qquad L_y=\inf\{t:y+W_t^2=1\}.
$$
On $\{H<L_y\}$ the path has stayed in the upper strip and first exits the cross-shaped domain at its right or left side. The boundary value is $a$ or $b$, respectively. Let $V=a$ if $x+W_H^1=1$ and $V=b$ if $x+W_H^1=-1$. The <Brownian exit from an interval> probabilities give
$$
\mathbb E V=a\frac{1+x}{2}+b\frac{1-x}{2}
=\frac{a+b}{2}+\frac{a-b}{2}x.
$$
Couple all heights $y$ using the same two Brownian paths. By part (a), $H<\infty$ almost surely, and continuity makes $\min_{0\leq t\leq H}W_t^2$ finite. Thus
$$
\mathbb P(L_y\leq H)
=\mathbb P\left(\min_{0\leq t\leq H}W_t^2\leq1-y\right)\longrightarrow0.
$$
Let $K=\max\{|a|,|b|,|c|,|d|\}$. On the complement of this exceptional event, $f(B_T)=V$; on the event itself, $|f(B_T)-V|\leq2K$. The representation in part (c) therefore yields
$$
\left|u(x,y)-\mathbb E V\right|\leq2K\,\mathbb P(L_y\leq H)\longrightarrow0.
$$
This proves the <harmonic limit along an infinite strip arm>:
$$
\boxed{\lim_{y\to\infty}u(x,y)=\frac{a+b}{2}+\frac{a-b}{2}x.}
$$
\b[Far up the arm, the bottom is reached with vanishing probability before the horizontal coordinate exits.] The limit is consequently determined by the two side values alone.
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