Why are single bonds weaker than double?

Simply from how they're constructed. Since a pure double bond consists of 1 #sigma# bond and 1 #pi# bond, it is one #\mathbf(pi)# bond's worth stronger than a single bond.

All pure single bonds consist of one #sigma# bond, i.e. due to one head-on orbital overlap.

Below is an example a constructive #2p_z-2p_z# head-on overlap that forms a #sigma_(2p_z)# molecular orbital, where electron density lies in the white bulged region---between the atoms.

A nice example is the #\mathbf(sigma)# bond in #\mathbf("Cl"-"Cl")#.

All pure double bonds consist of an additional #pi# bond, i.e. due to a sidelong orbital overlap.

Below is an example of the #2p_x-2p_x# constructive/bonding overlap, where electron density lies in the white bulged region (above the atoms).

An explicit #pi# bond example is the #\mathbf(pi)# bond in #\mathbf("O"="O")#, a product of either a #2p_x-2p_x# sidelong overlap, or a #2p_y-2p_y# sidelong overlap (but not both), that forms a #pi_(2p_(x"/"y))# orbital, depending on which pair overlaps.

This #pi# bond is made in addition to the #sigma# bond that was already made upon forming the first #"O"-"O"# bond.

Therefore, since a pure double bond consists of 1 #sigma# bond and 1 #pi# bond, it is one #\mathbf(pi)# bond's worth stronger than a single bond.