Suppose two black holes are observable by means of their lensing effect.
If these holes are orbiting around each other, then is there a minimum distance from each other to be still observable as separate black holes? Has this distance to do with the background of stars, the image of which they deform by lensing? Is there a minimum distance at all?
Gravitational lensing happens when a massive object passes (from our point of view) in front of a light source. The light rays bend and the image of the source gets distorted. A light ray that passes at a distance $r$ from a spherical object of mass $M$ is deflected by an angle
$$\theta = \frac{4GM}{rc^2}$$
Strong lensing for a single black hole
Consider a black hole that passes in front of a star. The maximal distortion of the image of the star is obtained when the star is directly behind the black hole. In that case instead of a point we see the star as a ring, called Einstein ring, with angular size given by
$$\theta_E = \sqrt{\frac{GM}{c^2}\frac{d_s-d_b}{d_sd_b}}$$
where $d_s$ is the distance from us to the star and $d_b$ is the distance from us to the black hole. If the star is twice as distant than the black hole, $d_s = 2d_b$ and
$$\theta_E = \sqrt{\frac{2GM}{c^2}\frac{1}{4d_b}} = \frac{1}{2}\sqrt{\frac{r_b}{d_b}} = \frac{1}{2} \sqrt{\frac{d_b}{r_b}} \theta_b$$
where $r_b$ is the Schwarzschild radius and $\theta_b = \frac{r_b}{d_b}$ is the angular radius of the black hole.
We may take $\theta_E$ as the characteristic angular size of a lensing event. If our telescope has an angular resolution $\theta_t < \theta_E$ then we are able to detect that the image of the star is deformed.
We could try to search for black holes by looking towards the centre of the Milky Way, because the stars are denser and a star-black hole conjunction is more probable. The Sun is about $d=8$ kpc from the centre of the MW, if we assume a stellar black hole distant $d/2 = 4$ kpc passes in front of a star, we may write
$$\theta_E \approx 0.0016{''} \times \left( \frac{d}{8 \text{kpc}}\right)^{-1/2} \left(\frac{M}{10M_{\odot}}\right)^{1/2}$$
Comparing this result with the angular resolution limit from Earth-based observations (seeing) $\approx 1 {''}$, or even with Hubble's angular resolution $\approx 0.05 {''}$, the observation of a distant black hole via strong lensing appears almost hopeless.
Microlensing for a single black hole
This is not the end of the story, tough, because spatially resolving the Einstein ring is not the only way to detect a lensing event. A gravitational lens not only deforms the shape of the background star, but also amplifies its brightness (just like a lens!). The if the angular separation between the star and the black hole is $\phi$, then the amplification factor will be
$$A(u) = \frac{u^2+2}{u\sqrt{u^2+4}};\ \ u = \frac{\phi}{\theta_E}$$
If a black hole happens to transit near a background star ($\phi \approx \theta_E$ or less), then we can detect a increase in the brightness of the star, even if we can't see any distortion in the shape of the star, which still looks like a fuzzy point to our telescope. The duration of the lensing effect depends on the time it takes for the black hole to transit through the distance $\theta_E$, which is typically on the order of some months.
A microlensing event is much easier to detect and many have been found during the searches for MACHOs. Unfortunately it is not easy to infer the mass of the lensing object, because it depends on distance and the speed of the object, which are usually unknown. Therefore, it is often difficult to say if a given microlensing event was due to a black hole, or a neutron star, or a brown dwarf ecc...
What about binary black holes?
The question was specifically about binary black holes. Everything above also applies to binary systems, but here a new characteristic scale enters into the problem, which is the projected orbital distance $a$ between the two black holes, along with its corresponding angular size
$$\theta_a = a/d_b \approx 0.0025 {''} \times \left(\frac{a}{10 \text{AU}}\right) \left( \frac{d_b}{4 \text{kpc}}\right)^{-1}$$
If $\theta_a \approx \theta_E$ the light curve of a star passing behind the binary system would appear significantly different than the one due to a normal microlensing event.
If $\theta_a \gg \theta_E$ a transiting star could intercept only one of the two black holes, making it impossible to identify the system as a binary.
Otherwise, a particularly far system with a close orbit could have $\theta_a \ll \theta_E$. In this case, unless the star passed really close to the black holes, we would detect a lensing event attributable to a single object with mass equal to the sum of the masses of the two black holes.
This is the closest thing to a "minimum distance" I could think of. Therefore, the minimum distance between the black holes to be able to tell it is a binary system is not an absolute value, but depends on the masses of the black holes, the distance of the system from Earth and the availability of background stars.
