closed in dual space E* implies closed in product space F^E

I am trying to show that if a set C is closed in the dual of E, E*, then it must be closed in F^E (where F = complex numbers or reals)

So far I've figured that E*is a subset of F^E since E* is the collection of continuous (bounded) linear functionals on E and F^E is all the linear functionals on E.

Thus, I have to show that E* is closed in F^E.

I have begun by supposing that there is a net of functions (fa)a in E* converging to a function f. I have tried to show that f will necessarily be bounded, and thus f lies in E* making E* closed.

I can't seem to get it.

if each fa (read as f sub alpha) is bounded, then there is an Ma > 0 such that for all x in E |fa(x)|<= Ma.

since (fa)a converges to f,

for all neighborhoods N of f, there is a b in the directed set such that for all a>=b, fa is in N.

I tried for a contradiction by supposing f isn't bounded, but I haven't really gotten anywhere.