isomorphism from well ordered set to ordinal 
Given $\langle X,<_{_X}\rangle$ well ordered set, show there exists unique ordinal $\alpha$ such that $\langle X,<_{_X}\rangle\cong \langle \alpha,\in\rangle$

Proving the uniqueness is easy, but I have a problem proving existence, instead of searching for $\alpha$ itself I think it will be easier to show that there exists $\gamma$(ordinal) such that there is isomorphism between an initial segment of $\gamma$ and $X$(as initial segment of ordinals are ordinals):
If there no such $\gamma$ we can conclude that for every ordinal $\alpha$ there is $X_\alpha\subsetneq X$ such that $\langle X_\alpha,<_{_X}\rangle\cong \langle \alpha,\in\rangle$ by theorem(The comparison theorem for well-orders). 
Now I am not sure if I can stop with:

We got that the set $X$ has subset as large as arbitrary ordinal, which implies that $Ord$ is a set, contradiction

Or do I need to really construct a surjective function from $X$ to $Ord$ and then by the axioms $Ord$ is a set or something alike.
 A: Here's an approach that, in a sense, combines both methods you were considering. The idea is to inductively embed ordinals into $(X, \le_X)$ until you can't any more, and then deduce that the ordinal you finished up with is the ordinal you seek.
First note that $(0, \in)$ embeds trivially into $(X, \le_X)$.
Suppose there is an embedding $i_{\alpha} : (\alpha, \in) \hookrightarrow (X, \le_X)$, and consider the set $X_{\alpha} = X \setminus i_{\alpha}[\alpha]$.


*

*If $X_{\alpha}$ is empty, then you can deduce that $i_{\alpha}$ is an isomorphism.

*If $X_{\alpha}$ is inhabited, then it has a $\le_X$-least element $x_{\alpha}$, and then there is an embedding $i_{\alpha+1} : (\alpha + 1, \in) \hookrightarrow (X, \le_X)$ which extends $i_{\alpha}$ and satisfies $i_{\alpha}(\alpha) = x_{\alpha}$.


Now suppose $\beta$ is a limit ordinal and that there are embeddings $i_{\alpha} : (\alpha, \in) \hookrightarrow (X, \le_X)$ for each $\alpha < \beta$ such that $i_{\alpha} \subseteq i_{\alpha'}$ for all $\alpha \le \alpha' < \beta$. Define $i_{\beta} = \bigcup_{\alpha < \beta} i_{\alpha}$ and note that $i_{\beta}$ is an embedding $(\beta, \in) \hookrightarrow (X, \le_X)$.
This process must eventually terminate, since $(\alpha, \in)$ does not embed into $(X, \le_X)$ whenever $|\alpha| > |X|$, and the only way it can terminate is when for some $\alpha$ we have $i_{\alpha} : (\alpha, \in) \hookrightarrow (X, \le_X)$ and $X_{\alpha} = \varnothing$.
There are a couple of gaps in the proof to be filled in, but I'm sure you can flesh out the details.
