
Uniformly continuous
Assume that (f_n) converges uniformly to f on A and that each f_n is uniformly continuous on A. Prove that f is uniformly continuous on A.
f_n converges uniformly so f_n(x)f(x)<epsilon.
f_n is uniformly continuous so xy<delta implies f_n(x)f_n(y)<epsilon

it's a standard trick (also known as "3epsilon trick"):
write f(x)f(y) < f(x)f_n(x) + f_n(x)f_n(y) + f_n(y)f(y) < epsilon+epsilon+epsilon
just be careful with the order of all those epsilons, deltas and n's

Oh, triangle inequality. Didn't think of that.