 Solution

Across a straight impermeable wall the reflected potential is obtained by conjugating the potential after reflecting its argument. For the real axis the operation is $f(z)\mapsto \overline{f(\bar{z})}$; the sum is real on that axis. For the imaginary axis it is $f(z)\mapsto \overline{f(-\bar{z})}$. Applying both commuting reflections to the dipole producesThe other three poles are outside the first quadrant. The sum is invariant under each reflected-conjugation operation, hence real on each boundary axis. Both walls are therefore streamlines.