I was working with the next exercise but I don't know how to procced. Any hint? The exercise comes from Rings of Continuous Functions by Leonard Gillman and Meyer Jerison.
$3$ J. Let $S$ denote the subspace of $\mathbb{R}\times\mathbb{R}$ obtained by deleting $(0,0)$ and all points $\left(1/n,y \right)$ with $y\neq 0$ and $n\in\mathbb{N}$. Define $\pi(x,y)=x$ for all $(x,y)\in S$; then $\pi$ is a continuous mapping of $S$ onto $\mathbb{R}$. Let $E$ denote the quotient space of $S$ associated with the mapping $\pi$; thus, $E$ may be identified as the set of real numbers endowed with the largest topology for wich the mapping $\pi$ is continuous (A set $A\subseteq E$ is open in $E$ if and only if $\pi^{-1}[A]$ is open in $S$.)
$1.$ $E$ is a Hausdorff space.
$2$. $E$ is not completely regular. [The set $\{ 1/n\}_{n\in\mathbb{N}}$ is closed]
$3$. $C(E)=C(\mathbb{R})$ where, in general, $C(X)=\left\{ f:X\to\mathbb{R}:f \ \text{is continuous} \right\}$
$4$ $E\cap[0,1]$ is pseudocompact, but not countably compact or completely regular.
I already have the solution of $1$, $2$ and the next partial solutions of $3$ and $4$.
For $3$.
If we denote by $\tau_s$ the topology of $E$ and $\tau_e$ the usual topology of $\mathbb{R}$ and we consider $\pi:\mathbb{R}^2\to\mathbb{R}$ then is continuous and $\pi_{|S}:S\to\mathbb{R}$ too. Then, the euclidean topology of $\mathbb{R}$ is a topology such that $\pi$ is continuos. Because $E$ have the largest topology for wich $\pi$ is continuous then $\tau_e\subseteq\tau_s$. Thus, if we consider $f\in C(\mathbb{R})$ then $f\in C(E)$. Therefore, $C(\mathbb{R})\subseteq C(E)$. For the another implication, Is what I did right?: If we take $f\in C(E)$ and an open set $U$ in $(\mathbb{R},\tau_e)$ then $f^{-1}[U]\in \tau_s$. We need to prove that $f^{-1}[U]\in\tau_e$. Because $f^{-1}[U]\in \tau_s$ then $\pi^{-1}\left[ f^{-1}[U]\right]$ is open in $S$. Therefore, there exist $V\subseteq\mathbb{R}^2$ an usual open set such that $\pi^{-1}\left[ f^{-1}[U]\right]=S\cap V$. Thus, $\pi\left[\pi^{-1}\left[ f^{-1}[U]\right] \right]=f^{-1}[U]=\pi[S\cap V]\subseteq \pi[S]\cap \pi[V]=\mathbb{R}\cap\pi[V]=\pi[V]$ and here I'm stuck...
For $4$.
Take $f:E\cap[0,1]\to\mathbb{R}$ continuous. By $3$, then $f:([0,1],\tau_e)\to\mathbb{R}$ is continuous and because $[0,1]$ is compact in $\tau_e$ is bounded.
$E\cap [0,1]$ is not countably compact because the set $\{1/n \}_{n\in\mathbb{N}}\subseteq E\cap [0,1]$ is a closed infinite discrete set.
I think that in $E\cap [0,1]$ the sets $\left\{ 1/n:n\in\mathbb{N}\right\}$ and $\{0\}$ can't be separated by a function, but, is it true? Here, again, I'm stuck.
I really appreciate any help you can provide.