Really there are two possible questions you might have. Why is an-am=(an-L)-(am-L)? And why is it useful to do this seemingly convoluted step?
As for the truth of the equality: since L-L=0, and adding 0 doesn't change anything, we have an-am=an-am+L-L. Now just group an-L and -am+L together to obtain that this is equal to (an-L)+(-am+L), which in turn is equal to (an-L)-(am-L).
Now for why this is useful: convergence of a sequence to L allows us to place neat bounds on terms of the form |an-L|. A sequence being Cauchy, on the other hand, means that terms of the form |an-am| can be bounded in a certain way. Now in this proof, we *know* that we can bound |an_L| and |am-L|, and we *want* to bound |an-am|. So our best bet is to bound |an-am| by the two terms we *can* bound. In order to do this, we need to introduce two copies of L, as is done above.