Let $F=\{p\in E\hspace{0.17em}|\hspace{0.17em}p\hspace{0.17em}\text{is an interior point of}\hspace{0.17em}E\}\text{.}$

To show: $E^{\circ }=F\text{.}$

To show:

(a) To show: If $p\in E^{\circ }$ then $p$ is an interior point of $E\text{.}$
Assume $p\in E^{\circ }\text{.}$
To show: $p$ is an interior point of $E\text{.}$
Since $E^{\circ }$ is open and $p\in E^{\circ }$ and $E^{\circ }\subseteq E,$ $p$ is an interior point of $E\text{.}$

(b) To show: If $p$ is an interior point of $E$ then $p\in E^{\circ }\text{.}$
Assume $p$ is an interior point of $E\text{.}$
To show: $p\in E^{\circ }\text{.}$
There is a neighbourhood $U$ of $p$ with $U\subseteq E\text{.}$
Since $U$ is open and $U\subseteq E$ then $U\subseteq E^{\circ }\text{.}$
So $p\in E^{\circ },$ because $p\in U\text{.}$

$\square$

To show:

(a) Assume $f$ is normal continuous.
To show: $f$ is topology continuous.
To show: If $V\subseteq ℝ$ is open then $f^{-1}\left(V\right)$ is open.
Assume $V\subseteq ℝ$ is open.
To show: $f^{-1}\left(V\right)$ is open.
To show: If $p\in f^{-1}\left(V\right)$ then $p$ is an interior point of $f^{-1}\left(V\right)\text{.}$
Assume $p\in f^{-1}\left(V\right)\text{.}$
To show: $p$ is an interior point of $f^{-1}\left(V\right)\text{.}$
We know $f\left(p\right)\in V\text{.}$
Since $V$ is open, $f\left(p\right)$ is an interior point of $V\text{.}$
So there exists $\epsilon \in {ℝ}_{>0}$ such that $B_{\epsilon }\left(f\left(p\right)\right)\subseteq V\text{.}$
Since $f$ is normal continuous there exists $\delta \in {ℝ}_{>0}$ such that if $d(x,p)<\delta$ then $d(f\left(x\right),f\left(p\right))<\epsilon \text{.}$
So there exists $\delta \in {ℝ}_{>0}$ such that $f\left(B_{\delta }\left(p\right)\right)\subseteq B_{\epsilon }\left(f\left(p\right)\right)\text{.}$
So $f\left(B_{\delta }\left(p\right)\right)\subseteq V\text{.}$
So $B_{\delta }\left(p\right)\subseteq f^{-1}\left(V\right)\text{.}$
So $p$ is an interior point of $f^{-1}\left(V\right)\text{.}$

(b) Assume $f$ is topology continuous.
To show: $f$ is normal continuous.
To show: If $p\in [a,b]$ and $\epsilon \in {ℝ}_{>0}$ then there exists $\delta \in {ℝ}_{>0}$ such that if $d(x,p)<\delta$ then $d(f\left(x\right),f\left(p\right))<\epsilon \text{.}$
Assume $p\in [a,b]$ and $\epsilon \in {ℝ}_{>0}\text{.}$
To show: There exists $\delta \in {ℝ}_{>0}$ such that $$f\left(B_{\delta }\left(p\right)\right)\subseteq B_{\epsilon }\left(f\left(p\right)\right)\text{.}$$ To show: There exists $\delta \in {ℝ}_{>0}$ such that $$B_{\delta }\left(p\right)\subseteq f^{-1}\left(B_{\epsilon }\left(f\left(p\right)\right)\right)\text{.}$$ Since $f$ is topology continuous and $B_{\epsilon }\left(f\left(p\right)\right)$ is open then $f^{-1}\left(B_{\epsilon }\left(f\left(p\right)\right)\right)$ is open.
So $p$ is an interior point of $f^{-1}\left(B_{\epsilon }\left(f\left(p\right)\right)\right)\text{.}$
So there exists $B_{\delta }\left(p\right)$ with $B_{\delta }\left(p\right)\subseteq f^{-1}\left(B_{\epsilon }\left(f\left(p\right)\right)\right)\text{.}$

$\square$