Let $\beta X$ denote the usual Stone-Čech compactification of a Tychonoff space $X$. Let $S \subset \beta\mathbb{R}$ denote those points in $\beta \mathbb{R}$ which can be obtained as a limit point of a sequence of points in $\mathbb{R}$ (since there are $2^c$-many points in $\beta\mathbb{R}$, $S$ seems small in size in $\beta\mathbb{R}$ in a certain sense).
My question is, is $S$ open in $\beta\mathbb{R}$? Seems a bit counter-intuitive, but I cannot prove or disprove this. Thanks in advance for your help!
Edit: In the first version of this question, I mistakenly asked about points which can be obtained as the limit of a sequence of real numbers, and as has been pointed out below, in that case $S = \mathbb{R}$.
The set $S$ is all of $\beta\mathbb{R}$. Indeed, let $(x_n)$ be any enumeration of $\mathbb{Q}$. Then $\{x_n\}$ is dense in $\mathbb{R}$, and hence also dense in $\beta\mathbb{R}$. So every point of $\beta\mathbb{R}$ is a limit point of the sequence $(x_n)$.
The following addresses the original version of the question, where $S$ was the set of limits of sequences in $\mathbb{R}$.
Suppose $x\in S\setminus\mathbb{R}$ and let $(x_n)$ be a sequence of distinct points in $\mathbb{R}$ converging to $x$. Then there is a continuous function $f:\mathbb{R}\to[0,1]$ such that $f(x_n)=0$ if $n$ is odd and $f(x_n)=1$ if $n$ is even (since $\{x_n\}$ is a discrete closed subset of $\mathbb{R}$). This $f$ cannot be continuously extended to $x$, which is a contradiction.
Thus $S=\mathbb{R}$. Since $\mathbb{R}$ is locally compact, it is open in $\beta\mathbb{R}$, so $S$ is open.