real analysis - Extending $f: Etomathbb{R}$ continuous to $g:mathbb{R}tomathbb{R}$ continuous

The following is from Carothers' Real Analysis:

Let $E\subset \mathbb{R}$ be closed and let $f:E\to\mathbb{R}$ be continuous. Prove that $f$ extends to a continuous function on all of $\mathbb{R}$. That is, prove that there exists a continuous function $g:\mathbb{R}\to\mathbb{R}$ such that $f(x)=g(x)$ for $x\in E$.

Thoughts regarding the proof:

Let $E\subset \mathbb{R}$ be closed and suppose $f:E\to\mathbb{R}$ is continuous. Since $E$ is closed is given, I suspect I will need to use a sequence of continuous functions which converges but I'm not sure how to incorporate this if this is indeed the case.

A hint to get me started on the proof would be appreciated. Thanks.

Answer

hints since $E$ is closed in $\mathbb{R}$ so $E^c$ is open in $\mathbb{R}$...now we know that every open set in $\mathbb{R}$ is countable union of disjoint open intervals...let {$(a_i,b_i)$} s.t $i \in \mathbb{N}$ is the set of all disjoint open interval...then for some $k \in \mathbb{N}$ we know the value of $f(a_k)$ and $f(b_k)$...so we can define a function $g_k$ in the interval $[a_k,b_k]$ s.t $f(a_k)=g(a_k)$ and $f(b_k)=g(b_k)$...and then then gluing up all such $g_k's$ and $f$ we can continuously extend the funtion...I think this much hint is enough..you can fill up all the details