By definition of accumulation point there is a sequence of distinct points .
But by the given you also know that .
If it were the case that then there is an open set containing and no point of .
There is a contradiction there.
Now since , we HAVE that for every +ve NO (and thus for a-L) ,THERE exists a δ>0 and such that:
if xεD and 0<|x-c|<δ,then |f(x)-L|<a-L.
But : |f(x)-L|<a-L <====> L-a<f(x)-L<a-L ====> f(x)<a ,a contradiction.
In the same way we work for b