How to show that $^{\omega}$ is not locally compact in the uniform topology?

On page 186 of Munkres’ Topology

Show that $[0,1]^{\omega}$ is not locally compact in the uniform topology?

Uniform topology is defined as topology induced by uniform metric $p$ which is stated as follows.

For any two points $a$, $b$ in $\mathbb{R}$, $$\bar{d}(a,b) = \text{min}\{|a-b|,1\}$$
For any two points $x$, $y$ in $[0,1]^{\omega}$ $$x = \{x_i:i<\omega\}$$ $$y=\{y_j:j<\omega\}$$ $$p(x,y) = \text{sup}\{\bar{d}(x_i,y_i):i<\omega\}$$

Is there some method to gain intuition on infinite product of topological space in various topologies? Could we visualize it as the finite product case?

Solutions Collecting From Web of "How to show that $^{\omega}$ is not locally compact in the uniform topology?"

Use the equivalent definition of local compactness: if $x \in U \subset C$, $U$ open, $C$ compact, then there must be a ball $B_{\epsilon}(x)$ whose closure $\bar{B}$ is contained in $U$. Apply this to $x = 0$. Then the closure of such a ball is compact as it is a closed subset of a compact set in a metric space. But it is not: look at the set $A = \{ x_i = (0, \dots, 0, \epsilon, 0, \dots ), i \in \mathbb{N} \} \subset \bar{B}$, with the $\epsilon$ in position $i$. It has no limit point $x$ in $A$ ($x$ cannot contain a coordinate in $(0, \epsilon)$ or a small enough ball around $x$ will not contain any $y \in A$; and in the other case, it is at distance $0$ or $\epsilon$ from any point in $A$ – think about that!). So $A$ is not limit-point compact, which in metric spaces is equivalent to being compact. Contradiction. 

(1). Generalities. For any space $X:$ If $Y$ is a closed discrete subspace of $X$ then every $Z\subset Y$ is closed: $\quad Cl_Y(Z)=Z$ because $Y$ is a discrete space, so $Cl_X(Z)=Cl_X(Z)\cap Cl_X(Y)=Cl_X(Z)\cap Y=Cl_Y(Z)=Z.$

Let $Y$ be an infinite closed discrete subspace of $X.$ Let $[Y]^{<\omega}$ be the set of finite subsets of $Y.$ (This is a standard notation in set-theory.) For each $W\in [Y]^{<\omega}$ the set $(X$ \ $Y)\cup W$ is open (…because its complement $Z=Y$ \ $W$ is closed). Now $\{(X$ \ $Y)\cup W: W\in [Y]^{<\omega}\}$ is an open cover of $X$ with no finite sub-cover, so $X$ is not compact.

(2). Specifics. Let $x=(x_j)_{j\in \omega}\in N\subset [0,1]^{\omega}$ where $N$ is a nbhd of $x. $ Take $r\in (0,1)$ such that the open ball $B_p(x,r)\subset N.$ For $n\in \omega$ let $y(n)=(y_{n,j})_{j\in \omega}$ where $y_{n,j}=x_j$ when $n\ne j$, and $y_{n,n}\in [0,1]$ such that $|y_{n,n}-x_n|=r/2.$

Then $Y=\{y(n): n\in \omega\}\subset N.$ And $Y$ is an infinite closed discrete subspace of $[0,1]^{\omega}$ because $p(y(m),y(n))=r/2$ whenever $m\ne n.$ So $Y$ is (a fortiori) an infinite closed discrete subspace of $\overline N,$ so $\overline N$ is not compact.

(3). Remarks. A non-compact metrizable space has an infinite closed discrete sub-space. But some other spaces (e.g. $\omega_1$ with the $\epsilon$-order topology) are non-compact and have no infinite closed discrete subspaces.

The set of all bounded real sequences with the metric $p((x_n)_n,(y_n)_n)=\sup_n|x_n-y_n|$ is called $l^{\infty}.$ (“El-infinity”). It is a complete metric space. The topology generated by $p$ is called the topology of uniform convergence: Members of $l^{\infty}$ are bounded functions from $\omega$ to $\mathbb R$, and a sequence $(f_n)_n$ of members of $l^{\infty}$ converges (with respect to the metric $p$) to $f\in l^{\infty}$ iff $(f_n)_n$ converges unformly to $f$ (as a sequence of functions).