Spelling Out the Details that $[0,1]^\omega$ is not Locally Compact I am having trouble understanding gnometorule's answer to this question:

Show that $[0,1]^{\omega}$ is not locally compact in the uniform topology, the uniform topology being induced by the metric $\displaystyle p(x,y) = \sup_{n \in \Bbb{N}} |x_n - y_n|$

Particularly, I am having trouble following the part that $A$ has no limit point. I don't following gnometorule's reasoning at all. Also, even if I did understand gnometorule's reasoning, gnometorule evidently has only shown that $A$ has no limit point in $A$---this is not what limit point compactness says, which I suspect gnometorule is appealing to to derive a contradiction. gnometorule has to show that $A$ has no limit point in $\overline{B} := \overline{B}(0,\epsilon)$. Hopefully someone can spell out the missing details in gnometorule's answer. 
Martin Sleziak suggested I use sequential compactness, but I am having trouble with this as well. Let $x_i = (y^{(i)}_n)$ be the vector with $y^{(i)}_n = \epsilon$ if $n=i$ and $0$ in any other case. Then $\{x_i\}$ is a sequence in the compact metric space $\overline{B}$ and therefore there exists a convergent subsequence $\{x_{i_k}\}$ that converges to $z=(z_1,z_2,...)$. I.e., for every $\epsilon > 0$, there exists an $N \in \Bbb{N}$ such that $|z_n - y_n^{(i_k)}| \le p(z,x_{i_k}) < \epsilon$ for every $i_k \ge N$. This in turn implies that for every $\epsilon > 0$, there exists an $N \in \Bbb{N}$ such that $|z_n - y_n^{(i_k)}| < \epsilon$ for every $n \ge N$, which holds for each $i_k$. In other words, $\displaystyle \lim_{n \to \infty} (z_n - y_n^{(i_k)}) = 0$ for each $i_k$, as a sequence in $\Bbb{R}$ with the standard topology. But $\lim_{n \to \infty} y_n^{(i_k)} =0 $ for each $i_k$ (again, as a sequence in $\Bbb{R}$), so that $\displaystyle \lim_{n \to \infty} z_n = 0$...

The indices and sequences of sequences is what is screwing with my mind. I am not sure what to do at this point. 
 A: I suspect "no limit point $x$ in $A$" was a typo for "no limit point $x$ not in $A$". In fact, $A$ has no limit point in the whole space $[0,1]^\omega$ other than the points in $A$.
Recall that $x_i$ is the point $(0,\dots,\epsilon,\dots,0)$, where the $\epsilon$ is in the $i^\text{th}$ position, and $A = \{x_i\mid i\in\omega\}$. Let $z$ be any point, and write $z(i)$ for the $i^{\text{th}}$ coordinate of $z$. We'll show that if $z\notin A$, then $z$ is not a limit point of $A$.
Suppose that for some $i$, $z(i) \neq 0$ and $z(i)\neq \epsilon$. Then $\rho(z,x_i)\geq|z(i)-\epsilon|>0$ and $\rho(z,x_j)\geq |z(i)|>0$ for all $j\neq i$. So $z$ is not a limit point of $A$, since there is no point of $A$ in a ball around $z$ of radius $\delta<\min(|z(i)-\epsilon|,|z(i)|)$.
So we're left to consider $z$ such that every coordinate of $z$ is either $0$ or $\epsilon$. If exactly one coordinate of $z$ is $\epsilon$, then $z\in A$. Otherwise, if all coordinates of $z$ are $0$, then for all $i$, $\rho(z,x_i) = \epsilon$ witnessed by coordinate $i$. And if at least two coordinates of $z$ are $\epsilon$, say coordinates $i$ and $j$, then $\rho(z,x_i) = \epsilon$ witnessed by coordinate $j$, and for all $k\neq i$, $\rho(z,x_k) = \epsilon$ witnessed by coordinate $i$. So unless $z\in A$, $z$ is not a limit point of $A$, since there is no point of $A$ in a ball around $z$ of radius $\epsilon/2$. 
A: (1). Let $(X,p)$ be a metric space. Let $T\subset X$ and $0<r\in \Bbb R$ such that $p(u,v)\geq r$ for all distinct $u,v\in T.$ Then $T$ is a discrete closed subspace of X.
(2).  If $S$ is a discrete closed subspace of a space $X$ then every $T\subset S$ is also a discrete closed subspace of $X.$
(3). A compact space $C$ cannot have an infinite discrete closed subspace $S.$ Otherwise let $S\supset T=\{u_n:n\in \Bbb N\}$ where $T$ is infinite. Then $B=\{(C\setminus T)\cup\{u_j:j<n\}: n\in \Bbb N\}$ is an open cover of $C$ with  no finite sub-cover. (By (2), members of $B$ are open in $C$ .)
(4). For  $(x_n)_{n\in \omega}=x\in [0,1]^{\omega}$ let $V$ be a nbhd of $x$. We show that $V$ is not compact. 
Let $0<s\in (0,1/2]$ such that $V\supset \{y:p(x,y)<s\}.$ Let $r=s/2.$ 
For $m,n \in \omega$ let $y_{n,m}=x_n$ if $n\ne m,$ and let $y_{n,n}\in [0,1]$ such that $|y_{n,n}-x_n|=r .$ Let $u_n=(y_{n,m})_{m\in \omega}.$ Then $T = \{u_n: n\in \omega\}$ is an infinite subset of $V.$ 
For $n\ne n'$ we have $p(u_n,u_{n'})\geq|y_{n,n}-y_{n',n}|=|y_{n,n}-x_n|=r.$  So  $T$ is an infinite discrete closed subspace of the space $V$ by (1). So by (3),  $V$ is not compact. 
Footnote: In (3) we could, instead, let $B=\{(C\setminus S)\cup A: A\in [S]^{<\omega}\},$ where $[S]^{<\omega}$ is the set of all finite subsets of $S.$
Footnote: In (4) we actually have $p(u_n,u_{n'})=r$ for $n\ne n'.$
