Integral of wedge product of 1-forms is less than norms of one forms on Riemann surface

Question

The statement:

Let $A$, $B$ be a real 1-form on Riemann Surface $X$. Then $$ \int_X |A\wedge B | \leq ||A||||B||. $$ I would like to prove this without using area form (or volume-form). I know how to show this when $A$ and $B$ are supported in one chart and in general case I can reduce this to the case where these form are compactly supported. So what is left is the compactly supported case. I suspect that partition of unity will be useful in the proof.

Here's source of the problem: http://wwwf.imperial.ac.uk/~skdona/RSPREF.PDF p.74

Definitions:

Inner product for (1,0)-forms a and b: $$ = \int_X ia\wedge \overline{b} $$ Inner product for real 1-forms A and B: $$ = 2, $$ where $A^{1,0}$ denotes the (1,0)-component of $A$.

I copied the definitions from the book version of that pdf file and now I noticed that Donaldson uses different definition in his pdf-version for inner product of real 1-forms. Proofs of the Lemma are same in both versions.

Edit April 29th:

I ended up using volume-form for this problem. So unfortunately I can't edit my work anymore even If someone would give a good answer.