Let $\alpha$ be ordinal number.
I want to prove that $\alpha^+ = \alpha +1$.
Is it sufficient for me to follow below listed steps?
first, since $\alpha$ is a well-ordered set, we can define the summation of natural numbers.
second, then since $\alpha^+ = \alpha \cup \{\alpha\}$, it equals to $\alpha +1$
A set is a transitive set iff every member of it is a subset of it.
A set is an ordinal iff it is a transitive set which is well-ordered by $\in.$ A well-order on a set $S$ is an irreflexive linear order $<$ on $S$ such that whenever $\phi\ne T\subseteq S$ there is a unique $<$-least member of $T.$ The class of ordinals is usually denoted by $On.$ Some basic properties of ordinals:
(i). If $a, b\in On$ and $a\ne b$ then $a\in b$ or $b\in a.$ Note that it follows from $b$ being a transitive set that $a\in b\implies a\subset b.$
(ii). If $x\in y\in On$ then $x\in On.$ That is, the members of an ordinal (if any) are ordinals.
(iii). If $x\in On$ then $x\not \in x.$ Without the Axiom of Foundation (a.k.a. Regularity) we can consistently assume there exists a set $A$ with $A\in A,$ but such an $A $ cannot be an ordinal.
(iv). Any set of ordinals is linearly ordered by $\in.$
For your Q: If $a,b\in On$ and $a<b$ then $a\in b$ by def'n of $<$ for $On.$ Also $b$ is a transitive set so $a\in b\implies a\subset b.$ Hence $a\cup \{a\}\subset b.$
By (iv) we have $a\cup \{a\}\ne a.$ Hence by (i) we have $a<a+1$. We have $a<a+1=a\cup \{a\}\subset b.$ Therefore $a<a+1\leq b.$
REMARKS. We can define "finite ordinal" in several ways without the word "finite". E.g. $x\in On$ and $x$ is doubly well-ordered: Every non-empty $T\subset x$ has a $\in$-largest member as well as a $\in$-least member..... With the Axiom of Infinity we can show there exists the set $\omega\;$ ("\omega") of all finite ordinals, which is also the least infinite ordinal.