My attempt: A is well ordered, hence every subset of A has a minimal element;
Since ⟨A,≤1⟩ is WO, it is isomorphic to an ordinal β, which is finite. But α is an ordinal as well. So we can define a function g: β-> α;
If we define the function f to map the min element a1 of an arbitrary subset of A,say A1, to f(a1), then for every a' in A1 we have that a1<=a', because of the order preserving.
Also f(a1)<=f(a'). If β and α are order isomorphic, then for every b in A (or after my assumption we can say β as well) we have f(b)=g(b).
And since A has an initial element a1, then this a1 will satisfy f(a1)<g(a1).
Any help is appreciated!
Either I'm misreading the question or the following is a counterexample. Let $A=\omega$ with the usual ordering and let $\alpha=2$. For each natural number $n$, define $f_n:\omega\to2$ by $$ f_n(k)=\begin{cases}0&\text{if }k<n\\1&\text{if }k\geq n. \end{cases} $$ Then $f_0\succ f_1\succ f_2\succ\dots$, so $\preceq$ is not a well-ordering.