Sigma bonds, pi bonds and the puzzling symmetry of methane
I am going to attempt to answer this, based on the link supplied by @Kartik and expanding on @Mitheron’s comment above.
As we know, covalent bonding is the process of “filling up” partially filled outer orbitals by sharing electrons between atoms. Carbon’s structure [He]2s22p2, or, more accurately, [He]2s22px12py1. When bonding occurs, however, two things happen:
1.  One of the 2s electrons gets “promoted” to a p orbital. This is an “excited” state in which the electrons have slightly more energy (though not enough to remove any of them). The configuration therefore becomes [He]2s12px12py12pz1, and voila, we now have 4 unpaired electrons available to bond with other atoms.
2.  The extra energy allows the orbitals to “hybridize”. In effect, the 2s and 2p orbitals morph into 4 identical orbitals of slightly higher average energy. These orbitals are called sp3 orbitals (one part s, three parts p). The electronic structure of the ready-to-bond atom is now [He] (sp3)4.
These four sp3 orbitals of course repel each other equally, and hence end up pointing towards the vertices of an imaginary tetrahedron. Hence the regular structure of CH4 and analogous compounds.
Returning to the question, the sigma bonds formed by carbon are neither from overlapping s or overlapping p orbitals, but from overlapping sp3 orbitals. 
The thing that makes a bond a sigma bond is not the type of subshell from which the electrons “originated”. It is the fact that the orbital overlap directly, with the two atomic nuclei and the main concentration of the shared electron cloud in a straight line. Pi bonds, on the other hand, overlap “sideways”, with the main concentrations of the bonding electron cloud off to the sides of the bonded atoms.
Please chime in, people, if I have omitted or misunderstood anything.