Ok, ok, I'm understanding.
Fix E>0 and let xc[x-1,x+1]. Since fn -> f uniformly on [x-1,x+1], for some n, p(fn(z), f(z)) < E/3 for all zc[x-1,x+1]. This fn is continuous at x, so for some delta>0, d(x,y)<delta implies p(fn(x),fn(y))<E/3. Then for such y, d(x,y)<delta, p(f(x)...