I have a little problem here. How would I go about to prove that any planar graph which does not contain a diamond (let a diamond be a graph consisting of two triangles intersecting in one edge) is 4-colorable? Without using the 4-color theorem. Any advice as to the approach to the solution would be greatly appreciated.

hmm, seems like you can just say since there are no 'diamonds', there certainly aren't any K5's (complete graphs on 5 vertices), since the diamond you've described is a subgraph of the K5. Since you never have 5 vertices that are all adjacent to each other, you'd never need 5 colours

In fact, you'd never even need 4 colours. With no diamonds, the graph should be 3-colorable, because the diamond thing is a subgraph of a K4 as well

I assume that, by diamond, you mean something like this: (X's are vertices)

'''''X
''''/''\
''/'''''\
X-----X
''\''''''/
''''\''/
'''''X

(ignore all the ''' ...that's just for formatting)