Why is $^\mathbb{N}$ not countably compact with the uniform topology?

My question is: Why is $[0,1]^\mathbb{N}$ not countably compact with the uniform topology? How do you prove this? Do you use the countable open covering or do you use the accumulation point definition?

Also I tried to show that $\beta\mathbb{N}$ has no non constant converging sequences but ran into some trouble. How do you show this?

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$\newcommand{\cl}{\operatorname{cl}}$Since the first question is either misstated or based on a misunderstanding, I’ll answer the second.

Suppose that $\sigma=\langle p_n:n\in\Bbb N\rangle$ is a non-trivial sequence in $\beta\Bbb N$ converging to some $p\in\beta\Bbb N$. Then $\sigma$ is not eventually constant, and we may without loss of generality assume that it is one-to-one and that $p_n\ne p$ for all $n\in\Bbb N$. It follows that $D=\{p_n:n\in\Bbb N\}$ is a discrete set in $\beta\Bbb N$, so for $n\in\Bbb N$ there are pairwise disjoint clopen sets $U_n$ such that $p_n\in U_n$; let $\mathscr{U}=\{U_n:n\in\Bbb N\}$.

Define $f:D\to[0,1]$ by setting $f(n)=0$ if $n$ is even, and $f(n)=1$ if $n$ is odd; $D$ is discrete, so $f$ is continuous.

$$\bar f:\Bbb N\to[0,1]:n\mapsto\begin{cases}
f(p_k),&\text{if }n\in\Bbb N\cap U_k\\\\
0,&\text{if }n\in\Bbb N\setminus\bigcup\mathscr{U}\;.
\end{cases}$$

Let $F$ be the extension of $\bar f$ to $\beta\Bbb N$. Each $U_n$ is clopen in $\beta\Bbb N$, so $\cl_{\beta\Bbb N}(\Bbb N\cap U_n)=\cl_{\beta\Bbb N}U_n=U_n$, and it follows that $F(p_n)=f(p_n)$. That is, $F\upharpoonright D=f$. And $p\in\cl_{\beta\Bbb N}D$, so $$F(p)=\lim_{n\to\infty}f(a_n)\;,$$ which is absurd, since the limit obviously does not exist.

Essentially the same argument can be used to prove considerably more. Replace the function $f$ by any function from $D$ to $[0,1]$, and define $\bar f$ and $F$ as above; it’s still true that $F\upharpoonright D=f$. Let $K=\cl_{\beta\Bbb N}D$, and let $\overline F=F\upharpoonright K$; then $K$ is a compactification of $D$, and $\overline F$ is a continuous extension of $f$ to $K$. In other words, every continuous function from $D$ to $[0,1]$ has a continuous extension to the compactification $K$ of $D$, so $K$ is homeomorphic to $\beta D$. But $D$ is homeomorphic to $\Bbb N$, so $K$ is homeomorphic to $\beta\Bbb N$ and therefore has cardinality $2^{2^\omega}=2^{\mathfrak c}$. In particular, $K\setminus D$ cannot be a single point, and therefore $\sigma$ cannot be a convergent sequence.

Added: Here’s a hint for the corrected first question. For each $n\in\Bbb N$ define a point $x^n=\langle x^n_k:k\in\Bbb N\rangle\in[0,1]^{\Bbb N}$ by

$$x^n_k=\begin{cases}1,&\text{if }k=n\\0,&\text{if }k\ne n\;.\end{cases}$$

Let $D=\{x^n:n\in\Bbb N\}$, and show that $D$ is a closed, discrete set in $[0,1]^{\Bbb N}$ with the uniform topology. (This is very straightforward.)

Now for $n\in\Bbb N$ let $D_n=D\setminus\{x^n\}$, let $V_n=[0,1]^{\Bbb N}\setminus D_n$, let $\mathscr{V}=\{V_n:n\in\Bbb N\}$, and show that $\mathscr{V}$ is a countable open cover of $[0,1]^{\Bbb N}$ with no finite subcover. (In fact no proper subfamily of $\mathscr{V}$, finite or infinite, covers $[0,1]^{\Bbb N}$: for each $n\in\Bbb N$, $V_n$ is the only member of $\mathscr{V}$ that contains $x^n$.)