Remainder of Stone-Čech is not first countable I have some troubles with the proof of the space $\beta X\setminus X$ is not first countable, where $\beta X$ is the Stone-Čech compactification. I have a little hint, but, sadly, I'm so stuck also with the hint. I need help.
I'm doing the proof by contradiction. Then, let  $p\in\beta X\setminus X$ and $\{G_n:n\in\mathbb{N}\}$ a local basis of $p$ composed by open sets. Then, because Stone-Čech is $T_4$, particularly, is $T_1$ and then, $\{p\}=\displaystyle\bigcap_{n\in\mathbb{N}} G_n$, i.e., $\{p\}$ is $G_\delta$. Then, we have to prove the next steps to conclude the proof.
1) $\{p\}$ is a zero set.
For each $n\in\mathbb{N}$, take $y_n\in\beta X\setminus G_n$. Because $\beta X$ is Tychonoff, there exists a continuous function $f_n:\beta X\rightarrow \left[0,\frac{1}{2^n}\right]$ such that $f_n(p)=0$ and $f_n(y_n)$. Then, let $f:\beta X\to \mathbb{R}$ defined by $f(x)=\displaystyle\sum_{n=1}^{\infty} f_n(x)$. Clearly $f$ is continuous and $f^1[\{0\}]=\{p\}$
2) $0\in\text{cl}_{\mathbb{R}}(f[X])\setminus f[X]$, where $\text{cl}_X(A)$ is the closure of $A$ in the space $X$.
Clearly, $0\notin f[X]$. Now, let $V$ an open neighborhood of $0$. If $V\cap f[X]=\emptyset$, then, $f^{-1}[V]\cap X=\emptyset$ but this is impossible, because $X$ is dense in $\beta X$ and $f^{-1}[V]$ is a non empty open set. Thus, $0\in\text{cl}_{\mathbb{R}}(f[X])\setminus f[X]$.
3) There exist two disjoint closed sets A and B in $f[X]$ such that $0\in\text{cl}_{\mathbb{R}}(A)\cap\text{cl}_{\mathbb{R}}(B)$
Take two disjoint sequences $\{a_n\}$ and $\{b_n\}$ that converges to $0$. They are closed in the space $f[X]$ and $0\in\text{cl}_{\mathbb{R}}(\{a_n\})\cap\text{cl}_{\mathbb{R}}(\{b_n\})$
4) Let $g=f_{|X}:X\to\mathbb{R}$. Then, prove that $\text{cl}_{\beta X} (g^{-1}[A])\cap\text{cl}_{\beta X} (g^{-1}[B])=\emptyset$
5) Prove that $\text{cl}_{\beta X} (g^{-1}[A])\cap\text{cl}_{\beta X} (g^{-1}[B])\neq\emptyset$
I'm so stuck with 4) and 5) steps. Clearly, from 4) and 5) follows the contradiction, but, really I don't know what can I do.
 A: Something much stronger is true, see Gilman and Jerrison (rings of continuous functions, theorem 9.5):

If $Z$ is a non-empty  closed $G_\delta$-set in $\beta X$ and $Z \cap X = \emptyset$, then $Z$ contains a homeomorphic copy of $\beta \mathbb{N}$, and so $|Z| \ge 2^{|\mathbb{R}|}$.

So in particular singletons cannot be $G_\delta$.
As to your proof: for 4 I think we can show that $p$ is the intersection of closures? Not quite clear. Where does this proof idea come from?
A: For 4.- We have that $f[X]$ is normal so there exists $h: f[X] \rightarrow [0,1]$ that satisfies 
$h[A]\subseteq \lbrace 0 \rbrace $ and $h[B]\subseteq \lbrace 1 \rbrace $ Then 
$h\circ g : X \to [0,1]$ can be extended to $\beta X$ call $\phi$ that extention, notice that $h\circ g [g^{-1}[A]]\subseteq \lbrace0\rbrace$  and $h\circ g [g^{-1}[B]]\subseteq\lbrace 1 \rbrace$ use this fact and $\phi$ to take two disjoint closed sets in $\beta X$ that separates $g^{-1}[A]$ and $g^{-1}[B]$.
For 5.- From 4 $\text{cl}_{\beta X} (g^{-1}[A])\cap\text{cl}_{\beta X} (g^{-1}[B])=\emptyset$ in particular $p\notin\text{cl}_{\beta X} (g^{-1}[A])\cap\text{cl}_{\beta X} (g^{-1}[B])\neq\emptyset$ supose wlg $p\notin \text{cl}_{\beta X} (g^{-1}[A])$ so there exists $U\subseteq\beta X$ open that satisfies $\text{cl}_{\beta X} (g^{-1}[A])\subseteq\beta X \setminus U$ send this with $f$ and see what happens. (Remember the fact that $f$ is perfect). 
