Following the previous question: Closure of a set of real-valued functions… Following the previous question:
Let $\mathcal F(\mathbb R)$ be set all of real valued function on $\mathbb{R}$ and $S\subset \mathcal F(\mathbb R)$ such that $f\in S$ if only if there is an interval $I$ and a polynomial $p\in \mathbb{R}[x]$ such that
$$f(x)=p(x)$$
for all $x\in I$. Now consider the set $\bar{S}$, where $f\in \bar{S}$ if only if there is sequence of function $(f_n)_{n=1}^{\infty}$ in $S$ such that 
$$f_n(x)\rightarrow f(x)$$ 
pointwise convergence for any $x\in \mathbb{R}$. My question is this:
$$C(\mathbb{R})\subset \bar{S}?$$
where $C(\mathbb{R})\subset \mathcal F(\mathbb R)$ is set of all continuous functions. If it is not true, is there any counterexample?  
@TZakrevskiy: according his answer $C(\mathbb{R})\subset \bar{S}$. My second question is this:
$$\bar{S}=\mathcal F(\mathbb R)?$$
If it is not true, is there any counterexample?
Is the following argument true? 
Let $g$ be any function and $$f_n(x)=\begin{cases}x+g(0),&x\in I_n,\\g(x),&x\notin I_n,\end{cases}$$  in which $I_n=[0,1/n]$, then $f_n$ pointwise to $g$ on $\mathbb{R}$! Therefore $\bar{S}=\mathcal F(\mathbb R)$.
It seems right. So I redefine my question as follows: 
$S\subset \mathcal F(\mathbb R)$ such that $f\in S$ if only if there is an unbounded interval $I$ and a polynomial $p\in \mathbb{R}[x]$ such that
$$f(x)=p(x)$$
for all $x\in I$. In this case:
 $$C(\mathbb{R})\subsetneqq \bar{S}‎\subsetneqq‎ \mathcal F(\mathbb R)?$$ 
 A: Your argument works. You could simplify it a bit by making $f_n(x) = g(0), x \in I_n$ (remember that constant functions are also polynomials), but that isn't necessary. As for your redefined question, use $I_n = [n, \infty)$, and it isn't hard to see that $\bar{S} = \mathcal{F}(\Bbb R)$ still holds.
Unless by "unbounded interval", you meant unbounded on both sides. Then the answer is different because the only such interval is $(-\infty, \infty)$, which means $S = \Bbb R[x]$.
A: I only answer the case where in the definition of $S$, the interval $I$ is always $(-\infty, \infty)$ (The other cases has been answered in another answer, if in the definition of $S$ you mean $I$ is depending on $f\in S$)
In this case $S = \mathbb R[x]$. I claim that 
$$C(\mathbb{R})\subsetneqq\bar{S}‎\subsetneqq‎ \mathcal F(\mathbb R).$$
Claim one: $C(\mathbb R) \subset S$. Let $f\in C(\mathbb R)$. Then for each $n$, there is $p_n \in S$ so that 
$$|f(x) - p_n(x) | <\frac 1n$$
whenever $x\in [-n, n]$. Thus $p_n$ converges pointwisely to $f$. 
Claim two: $C(\mathbb R) \neq \overline S$. 
Let $f_n\in C(\mathbb R)$ so that 
$$f_n(x) =\begin{cases}
1 &\text{on }[-n, 0],\\ 
0 & \text{on }[\frac 1n, n],\\
 \text{linear} &\text{on } [0, \frac 1n].\end{cases}$$
Let $p_n \in S$ so that 
$$|f(x) - p_n(x)|<\frac 1n$$
on $[-n, n]$. Then $p_n$ converges pointwisely to 
$$f(x) = \begin{cases} 1 & \text{if } x\le 0 \\ 0 &\text{if }x>0.\end{cases}$$
Claim three: $\overline S \neq \mathcal F(\mathbb R)$. 
(Might be a overkill) As pointwise limit of measurable functions are still measureable, $\overline S$ is contained in the set of all measurable function, thus is not $\mathcal F(\mathbb R)$. 
