For $y\in Y$ let $(-,y)=\{y'\in Y: y'<y\}$ and $(y,+)=\{y'\in y:y<y'\}.$
Let $L=\{(-,y):y\in Y\}$ and $U=\{(y,+):y \in Y\}.$
Let $B=\{Y\} \cup L \cup U \cup \{l\cap u:l\in L\land u\in U\}.$
Then $B$ is a base for the order-topology on $Y. $
For $y,y'\in Y$ let $(y,y')=\{y''\in Y: y<y''<y'\}.$ This is called a bounded open interval. The members of $\{l\cap u:l\in L\land u\in U\}$ are the bounded open intervals (including the empty set).
Any $A\subset Y$ is convex iff $\forall a,a'\in A\; (\,(a,a')\subset A).$The members of the base $B$ are precisely the convex open sets in the order-topology. So any open set in the order-topology is the union of a family of convex open sets.
This is applicable to all $Y,$ i.e. regardless of whether $Y$ has a max or min or any other specific properties. In particular (as a caution) it is possible for a bounded open interval to have a max or min.
A useful property for any $Y$ is that the closure of $(-,y)$ is $(-,y)\cup \{y\}$ iff $y=\sup (-,y)$ iff $(-,y)$ is not empty and has no max. And $(-,y)$ is closed iff $(-,y)$ is empty or has a largest member. And similar considerations hold for $(y,+).$
From that, we see that if $y\in Y$ and $y\not \in A\subset Y$ then $y\in \overline A$ iff $[\,y=\sup (A\cap (-,y)) \lor y=\inf (A\cap (y,+)\,)\,].$
To show that $f$ is continuos, we have to show that for all $A$ open set of $\mathbb N$, $f^{-1}(A)$ is an open set. An open set of $\mathbb N$ with the co-finite topology is a set such that $\mathbb N \ \backslash A$ is finite. Hence we have to show that $f^{-1}(A)$ is a set such that $\mathbb N \ \backslash \ f^{-1}(A)$ is finite.
Now:
$$\mathbb N \ \backslash \ f^{-1}(A) = \{n\in \mathbb N \ | \ n^3\notin A\} = f^{-1}(\mathbb N\ \backslash \ A)$$
By hypotesis $\mathbb N \ \backslash A$ is finite, and the map $f$ is injective; hence $\mathbb N \ \backslash \ f^{-1}(A)$ is finite.
Best Answer
I would define it as follows. A function $f:X\to Y$, where $X$ is an ordered set with the order topology and $Y$ is a topological space is continuous from the right at a point $x\in X$ if $x$ is the largest element of the set $X$ or, otherwise, for each neighborhood $U$ of the point $f(x)$ there exists a point $y\in X$ such that $y>x$ and $f([x,y))\subset U$.