Hmm, it might be possible...but unlikely.
To respond to the clarified question (can be found in the comments), let's assume that by "take up the sky" you mean that each planet has almost a 90 degree angular diameter when viewed from the other planet. This will take up 50% of the night's sky. (Getting 100% is impossible due to the curvature of a spherical object)
I have calculated that the distance necessary between two Earth-sized objects for the above scenario is about 15380 km. This leaves only 2638 km of distance between the outer surfaces of each planet! Uh-oh! That might be too close for each planet to be habitable!
With each planet taking up 25% of the sky, this distance increases to 23019 km, leaving around 10277 km of space between the surfaces of the planets. This is still extremely close.
However, you'd be surprised at how big objects will look, even if they do not take up most of the visable sky. For example, with an angular diameter of 5 (about 2.7% of the sky), the planets would appear to be about 10 times bigger in apparent size than the real-life full moon. The distance required for this sight is around 152429 km (about half the Earth-moon distance), which is a significantly greater distance and would almost certainly be habitable.
If you want to calculate the distance necessary for two Earth-sized objects, use the following formula, which I derived from the formula for angular diameter:
$$D =\frac{d}{2\;sin\left( \frac{\pi a}{360} \right)}$$
where $D$ is the distance between the two objects, $d$ is the actual diameter of the two objects (assuming they are identical), and $a$ is the desired angular diameter, measured in degrees.
The result (distance) of this formula will always be in the same units that you provided for the actual diameter.
Note: this part of my answer was written before the original poster clarified his question in the comments. It has been kept here because it offers Roche limit calculations:
What you are looking for is the Roche limit of both planets in your system.
Let us assume you have 2 exact replicas of the Earth:
For a rigid-satellite calculation, the Roche limit is determined by the ratio of the densities of the two objects, and is given by the following formula:
$$d=R_M \sqrt[3]{2\frac{p_M}{p_m}}$$
where $R_M$ is the radius of the primary (larger/more massive) body, $p_m$ is the density of the primary body, and $p_M$ is the density of the satellite.
Since $p_M$ and $p_m$ are the same, they cancel out and become $1$. The simplified equation then becomes:
$$d = R_M \sqrt[3]{2} $$
According to this simplified equation (it does not consider inertia force and rigid structure), there are not really limitations on the distance. This doesn't seem correct to me, but it might be due to the fact they are of comparable sizes so the force of one body does not dominate the other.