For a single policeman, $E\tau_1$ is finite if he has a probability ($1/2+\epsilon$) of moving towards the thief.

If there is a second policeman, further away from the thief than the first, both back on $p=\frac{1}{2}$ then there is a non-zero probability that he will catch up with the first in a finite time (provided that this time is sufficient, of course). When they move from the same position, there is a higher probability that one of them will move towards the thief. So, very informally, can we say that a second policeman effectively increases the probability of one policeman moving towards the thief? This gives us an equivalent case to my first paragraph, with a finite $E\tau$.

Here is a vague justification for the single-policeman case with probability $p$ of moving towards thief. Actually, it helped me to start with $p=\frac{1}{2}$: for that case, define $E_n$ as the expected number of further steps if the current separation is $2n$.

$E_n = 1+\frac{1}{4}E_{n-1}+\frac{1}{2}E_{n}+\frac{1}{4}E_{n+1}$

Rearrange:
$E_n = 2+\frac{1}{2}E_{n-1}+\frac{1}{2}E_{n+1}$

Apply this to both $E$'s on the RHS and rearrange:
$E_n = 8+\frac{1}{2}E_{n-2}+\frac{1}{2}E_{n+2}$

This formula can then be applied to itself in a similar way, and so on:

$E_n = 32+\frac{1}{2}E_{n-4}+\frac{1}{2}E_{n+4} = 128+\frac{1}{2}E_{n-8}+\frac{1}{2}E_{n+8}$
$E_n = 2^{2k+1}+\frac{1}{2}E_{n-2^k}+\frac{1}{2}E_{n+2^k}$

All of these are valid so long as there are no negative subscripts. However, we can use $E_0=0$:

$E_2=8+\frac{1}{2}E_4=8+\frac{1}{2}(32+\frac{1}{2}E_8)=8+16+32+\dots$

which is infinite as expected. But if we follow the same steps with probability $p>0.5$, the behaviour is different:

$E'_n = C_1+p_1 E'_{n-2}+(1-p_1) E'_{n+2}$

where $C_1=4/[1-2p(1-p)]$ and, more importantly, $p_1=p^2/[1-2p(1-p)]\approx (\frac{1}{2}+2\epsilon)>p$, so subsequent steps are increasingly different. Quite soon:

$\{E'\}_n \approx D 2^k +\{E'\}_{n-2^k} $ and so
$\{E'\}_{2^k} \approx D 2^k$.