Exercise 1, Section 18 of Munkres’ Topology 
Prove that for function $f:\Bbb{R} \to \Bbb{R}$, the $\epsilon - \delta$ definition of continuity implies the open set definition.

My attempt: Claim: The notion of open set in metric space $(\Bbb{R},d=\| \cdot \|)$ setting is same as the notion of open set in standard topology, $\mathcal{T}$, on $\Bbb{R}$. Proof: let $A\subset \Bbb{R}$ is open in metric space notion. Let $B=\{B(x,r)|x\in \Bbb{R},r\gt 0\}$. Then $A$ is open iff $A=\bigcup_{i\in I} B_i$, where $B_i\in B, \forall i\in I$ and $I$ is an indexing set(Proof: $A$ is open iff it is union of open balls). In $\Bbb{R}$, $B(x,r)=(x-r,x+r)$ is precisely segment in$\Bbb{R}$. So $B(x,r)\in \mathcal{B}$, $B\subseteq \mathcal{B}$. Since $B_i\in \mathcal{B}\subseteq \mathcal{T}, \forall i\in I$, we have $A=\bigcup_{i\in I}B_i \in \mathcal{T}$. Hence, $A$ is open with respect to standard topology $\mathcal{T}$. Conversely, suppose $A\in \mathcal{T}$. Then $A=\bigcup_{i\in I}(x_i, y_i)$, where $(x_i, y_i)\in \mathcal{B}, \forall i\in I$. Note $(x_i, y_i)=B(\frac{x_i+y_i}{2}, \frac{y_i-x_i}{2})=B_i$, $r=\frac{y_i-x_i}{2}\gt 0$, for seek of completeness, I compute $r$ by solving $\frac{x_i+y_i}{2}+r=y_i$ equation. Thus $A=\bigcup_{i\in I} B_i$ $\Rightarrow$ $A$ is open in $\Bbb{R}$, metric space notion. So we can interchange the notion of open set between metric space and topological space, provided $\Bbb{R}$ is equipped with standard topology and metric on $\Bbb{R}$ is the euclidean norm or absolute value map.
By theorem 4.8 of Baby Rudin, which states “A mapping $f$ of a metric space $X$ into a metric space $Y$ is continuous on $X$ if and only if $f^{-1}(V)$ is open $X$ for every open set in $Y$”. In our case, $X=Y=(\Bbb{R},\| \cdot \|)$. So we can write, $f$ is continuous by $\epsilon -\delta$ definition $\Leftrightarrow$ $f^{-1}(V)\in \mathcal{T}$, for every  $V\in \mathcal{T}$. Is this proof correct?

Application: We can use this result to prove following problems:
(1) Exercise $18.5$. Proof: Prove that $(a,b)$ is homeomorphic to $(0,1)$. $f:(a,b)\to (0,1)$, defined by $f(x)= \frac{x-a}{b-a}$ and $g:(0,1)\to (a,b)$, defined by $g(y)=(b-a)y+a$. By theorem 4.4 of Baby Rudin, $f$ & $g$ are continuous map, derived from $\epsilon -\delta$ definition of continuity, hence $f$ & $g$ are continuous maps in topological notion. By definition of map $g$, $\forall y\in (0,1)=Y, \exists !x\in (a,b)=X$ such that $g(y)=x=g(f(x))$. It is easy to check map $g$ is injective. So $g(y)=g(f(x))\Rightarrow y=f(x)$, which is exactly the definition of bijective map of $f$. Is this proof correct?
(2) Exercise $18.6$. Proof: it is easy to check $f(x)=\begin{cases}  0   &\text{ if } x\in \Bbb{Q} \\ x  &\text{ if } x\in \Bbb{R}\setminus \Bbb{Q} \end{cases}$ is continuous only at $x=0$ by $\epsilon -\delta$ definition.
(3) Exercise $18.12$ (a) and (b). Proof: fix $y_0 \in \Bbb{R}$. $h$ is continuous, by theorem 4.4 of Baby Rudin. By exercise 1, $h$ is continuous in open set definition. Similarly $k$ is continuous. Hence $F$ is continuous in each variable. $g: \Bbb{R}\to \Bbb{R}$ defined by $g(x)=F(x\times x)$ is continuous at $\Bbb{R}\setminus \{0\}$. Is this proof correct?
 A: Your whole attempt for the question in the title is: first show (in a way too complicated manner) that $\mathcal{T}_< = \mathcal{T}_d$ for $\Bbb R$ in the standard metric $d(x,y)=|x-y$ and standard order $<$.
Finally call upon a result in a totally different book (Rudin, instead of Munkres).
I'd say that's not a valid proof in a context where you're following Munkres' text and doing his excercises, sorry.
Munkres himself mentions in the text already that these ways of defining the topology on $\Bbb R$ yield the same result. (Example 2 in section 20).
21.1 in section 21 details why $\varepsilon$-$\delta$ continuity is the same (between metric spaces) as the topological definition. All of this is irrelevant to section 18 and its exercises.
Of course ex. 18.1 comes too early in the text in that sense.
But a direct proof is very possible: let $f: \Bbb R \to \Bbb R$ obey e-d continuity and let $O \subseteq \Bbb R$ be open. Then fix $x_0 \in f^{-1}[O]$. So $f(x_0) \in O$ and as $O$ is open we have some open interval $(c,d)$ so that $f(x_0) \in (c,d) \subseteq O$. Let $\varepsilon = \min(f(x-0)-c, d-f(x_0))$ then $B_\varepsilon(f(x_0)) \subseteq (c,d)$. We find a $\delta>0$ for the epsilon-delta continuity. It follows easily that $U:=(x-\delta, x+\delta)=B_\delta(x)$ obeys $f[U] \subseteq O$ or equivalently $U \subseteq f^{-1}[O]$. It follows that $f^{-1}[O]$ is open. The reverse implication is very similar.
For the "application" 18.5 it suffices to note that $f$ and $g$ are mutual inverses and increasing so homeomorphisms in the order topology. No metric needed at all.
18.6 does not need the epsilon-delta definition, the topological one will do fine.
For 18.12 you cannot use epsilon-delta arguments yet as they follow later, as said. For $y=0$ we just have the $\equiv 0$ function and for $y\neq0$ we get a quotient of continuous functions (linear and a quadratic) with a denominator that's never $0$. By old results/exercises this is continuous. Same for fixed $x$.
A: Let $f$ be continuous in the $\varepsilon-\delta$ sense, and let $V$ be an open subset of $\mathbb{R}$. Let $x_{0} \in f^{-1}(V)$. Then $V$ contains an interval of the form $\left(f\left(x_{0}\right)-\varepsilon, f\left(x_{0}\right)+\varepsilon\right)$ for some $\varepsilon>0$. By assumption, there exists $\delta>0$ such that $\left|x-x_{0}\right|<\delta$ implies $\left|f(x)-f\left(x_{0}\right)\right|<\varepsilon$. Thus $x \in\left(x_{0}-\delta, x_{0}+\delta\right) \subset f^{-1}(V)$ and therefore $f^{-1}(V)$ is open. It follows that $f$ is continuous in the open set definition.
