# $X=[0,1]^\omega= \prod_{n=1}^\infty [0,1]$ is compact.

I need to prove that $X=[0,1]^\omega= \prod_{n=1}^\infty [0,1]$ with product topology is compact without using Tychonoff theorem.(Here, $[0,1]$ with subespace topology of $\mathbb R$) Is $X=[0,1]^\omega$ compact with the uniform topology?

Show that the product topology on $[0,1]^\omega$ is metrisable and use that the product of countably many sequentially compact spaces is sequentially compact via the diagonal sequence technique.
In the uniform metric topology $[0,1]^\omega$ is not compact, as $\{e_1 = (1,0,0,0\ldots), e_2 = (0,1, 0, 0 , \ldots,), e_3=(0,0,1, 0, \ldots)\ldots\}$ is an infinite set that is closed and discrete.