Stone-Čech compactifications and limits of sequences I've been working on some old prelims from my university when they used to just be on point-set topology. We don't cover a couple of the topics so I've been teaching myself some of the material, one of the topics no longer covered is Stone-Čech Compactification. Which I have a somewhat tenuous understanding of, at any rate one of the questions reads:
Let $X$ be a completely regular topological space and let $\beta(X)$ denote the Stone-Čech compactification of $X$. Show that every $y \in \beta(X) \setminus X$ is a limit point of $X,$ but is not the limit of a sequence of points in $X$.
It's clear to me how to go about the first part, $X$ if considered as a subset of $\beta(X)$ is dense in $\beta(X)$. Then it follows that for every point $y \in \beta(X) \setminus X$ that every neighborhood of $y, U$ in $\beta(X)$ will touch $X$. 
For the second part of the question, I must say, sadly, that I'm at a loss in general. Presumably we need to assume that we have some convergent sequence $\{x_i\}_{i \in \mathbb{N}}$ that converges to a point $y \in \beta(X) \setminus X$ and show that this is a contradiction. But, probably due to my weak understanding of the Stone-Čech Compactification I am unsure of how to go about this.
 A: In Gillman & Jerison's Rings of continuous functions I found the
following note

8.21 N.B.  A number of authors have fallen into the trap of assuming
  then every countable, closed, discrete subset of a completely regular
  space is $C^\ast$-embedded. We have just seen a counterexample: [...]

It seems likely that one of these authors, or someone following them has
made up this problem, as the same counterexample works here. (Martin Sleziak's answer shows how the assumption would lead to a proof).
The counterexample referred to is the Tychonoff plank
$([0, \omega_1] \times [0, \omega]) \setminus (\omega_1, \omega)$.
It can be shown that every real-valued function on that space can
be extended to its one-point compactification 
$[0, \omega_1] \times [0, \omega]$.
That implies that this is also the Stone-Čech compactification, and
clearly $\lim_{n\to\infty} (\omega_1, n)  = (\omega_1, \omega)$.
A: EDIT: As pointed by Stefan H. in  his comment, the solution I have suggested only works if $X$ is normal, since I am using Tietze extension theorem.

Perhaps I have overlooked something and I will be blushing, but I will give it a try. (This is my solution, I did not check the books I mentioned in the above comment. Perhaps the proofs from those books can give you a hint for a different proof.)
Let $x_n\in X$ be a sequence which converges to $x\in\beta X\setminus X$. We can assume that $x_n$'s are distinct.
I will show bellow that $\{x_n; n\in\mathbb N\}$ is closed discrete subspace of $X$. But first I will show how to use this fact.
For any choice of $y_n\in[0,1]$, $n\in\mathbb N$, we can define $f(x_n)=y_n$ and extend it continuously (by Tietze's theorem) to the whole $X$. 
Now there exists a continuous extension $\overline f : \beta X \to [0,1]$. 
By continuity, the sequence $y_n=\overline f(x_n)$ converges to $\overline f(x)$. We have shown that every sequence in $[0,1]$ is convergent, a contradiction.

Now to the proof that  $\{x_n; n\in\mathbb N\}$ is closed and discrete.
Since $\{x_n; n\in\mathbb N\}\cup\{x\}$ is a compact subset of $\beta X$, it is closed in $\beta X$. 
The intersection with $X$ is $\{x_n; n\in\mathbb N\}$ and it must be closed in $X$.
Now we consider $\{x_n; n\in\mathbb N\}$ as a subspace of $X$ and we want show that this subspace is discrete.
Choose some $x_n$. By Hausdorffness, it can be separated from $x$, i.e. there exists a neighborhood $U\ni x$ such that $x_n\notin U$ and a
neighborhood $V\ni x_n$ with $V\cap U=\emptyset$.
Now by convergence $U$ contains all but finitely many $x_n$'s, hence using Hausdorfness we can separate $x_n$ from the (finitely many) remaining ones.
