Why are the supremums of this linearly interpolated function equal?

Original question:

Suppose $f$ is a real, continuous function defined on the closed set $E subset mathbb{R}^1$. Prove that there exists real, continuous function $g$ on $mathbb{R}^1$ such that $g(x) = f(x) forall x in E$.

Proof:

Suppose $f$ is a real, continuous function defined on the closed set $E subset mathbb{R}^1$. Then, $E^c$ is open in $mathbb{R}^1$. Then $E^c$ can be expressed as a union of an at most countable collection of pairwise disjoint open intervals and we can write $$E^c = cup_{i = 1}^{n} (a_i, b_i) cup (-infty, a_0) cup (b_0, +infty).$$ where either $n$ is finite or $n in mathbb{N}$ and $a_i < b_i$. Next, define $g$ as

$g(x) =
begin{cases}
f(x) & text{if $x in E$} \
f(a_i)+(x-a_i)frac{f(b_i)-f(a_i)}{b_i-a_i} & text{if $x in (a_i, b_i)$} \
f(b_0) & text{if $x in (b_0, +infty)$} \
f(a_0) & text{if $x in (-infty, a_0)$}
end{cases}$

Clearly, $g$ is an extension of $f$ on $mathbb{R}^1$ and it remains to show that $g$ is continuous on $mathbb{R}^1$.

Now, to show that $g$ is continuous if $xin E$, I found a proof that begins with:

Given $xin E$ and $epsilon>0$, by the continuity of $f$ on $E, exists delta_0>0$ such that $$yin E textrm{ and } |y-x|<delta_0 implies |f(y)-f(x)|<epsilon/2.$$ If there exists a $zin E$ with $x-delta_0<z<x$, set $delta_1=x-z$ and note that
begin{equation*}
sup_{x-delta_1<y<x} |g(y)-g(x)| = sup {|f(y)-f(x)|:yin E, x-delta_1<y<x} le epsilon/2 < epsilon
end{equation*}

My question: Why is it true that $sup_{x-delta_1<y<x} |g(y)-g(x)| = sup {|f(y)-f(x)|:yin E, x-delta_1<y<x}$? I think that this is true because $g$ is a linear extension of $f$. Why can the $sup$ be restricted to the $y$-values $x-delta_1<y<x$?