Q:

Show the existence of a sequence of polynomials that converges to $f$ on any compact subset of $\mathbb{R}$.

If $f$ is continuous function ($f:\mathbb{R}\rightarrow\mathbb{R}$), then there exists a sequence of polynomials which converges to $f$ on any compact subset of $\mathbb{R}$.
Proof:
(Weierstrass) If $f$ is a continuous complex function on $[a,b]$, then there exists a sequence of polynomials $P_n$ such that $$\lim_{n\rightarrow\infty}P_n(x)=f(x)$$ uniformly on $[a,b]$. 
Assume $K\subset\mathbb{R}$, compact, then it is certainly $K$ is bounded and closed. Since $K$ is bounded, $K$ (in real) contained in some interval. Let $A_n$ be the colosed interval $[-n,n]$, and let $\epsilon_n=1/n$. Then $\forall\epsilon_n>0$, by Weierstrass, there exists some sequence of polynomials $P_n(x)$ such that $\forall x\in A_n$, $|f(x)-P_n(x)|<\epsilon_n$. Using the fact that $$\lim_{n\rightarrow\infty}\epsilon_n=0$$ therefore, this constructed sequence of polynomials $P_n$, converges to $f$ on $K$.
Question: By such a construction, expanding intervals, are we taking subsequence of the previous sequence of polynomials?

A:

A small subtlety is that you should construct the sequence $\{P_n\}$ before you specify your compact set $K$. This will also have the added benefit that uniform convergence on $K$ is easier to see. Let's go through a little more slowly.
Since $[-n,n]$ is compact, there exists a sequence $\{Q_{n,k}\}$ of polynomials such that $Q_{n,k}\to f$ as $k\to\infty$ uniformly on $[-n,n]$. In particular, there exists $k_n$ such that $|Q_{n,k_n}(x)-f(x)|\le\frac1n$ for all $x\in[-n,n]$. Let $P_n:=Q_{n,k_n}$. This is our sequence of polynomials. So the short answer is yes: each interval $[-n,n]$ gives rise to a different sequence of polynomials, but we can choose a single sequence in a clever way which does not depend on the interval.
Now let $K\subset\mathbb R$ be an arbitrary compact set. Then $K\subseteq[-N,N]$ for some $N$. Thus, for every $n\ge N$, if $x\in K$ then $x\in[-n,n]$ and thus $|P_n(x)-f(x)|\le\frac1n$. Notice that $N$ was specified before $x\in K$, so $P_n\to f$ uniformly on $K$.

