Let $X$ be the set of all continuous functions from the unit interval $[0, 1]$ to itself.
(1) Prove that $X$ is not compact under the pointwise convergence topology.
(2) Prove that $X$ is compact under the uniform convergence topology.
To prove (1), I found the counterexample as the sequence of functions $\langle f_n \rangle$ defined as $f_n (x) = x^n$, which converges to $f(x) = 0$ for $0 \le x <1$ and $f(x)=1$ for $x=1$, which is not continuous, so $X$ is not sequentially closed. However, I remember that it cannot be guaranteed that sequential closedness does not imply closedness of topological space unless it is first countable, and $I^I$ under product topology is not first countable. How should I verify that $X$ is not compact?
Also, this counterexample can be applied in (2), which deduces that $X$ is not compact under the uniform convergence topology. Is there anything wrong in my argument?
As I know, in the metric space, uniform convergence topology is the induced topology from the uniform metric $d$ between two functions is given as $d(f, g) = \sup|f-g|$, right? Also, in the metric space, sequential compactness and compactness are equivalent. Is there any error in what I know?