I'm trying to show that for any countable ordinal $\alpha$, there is a subset of $(\mathbb{R},<)$ that has order type $\alpha$. In this post I'm not asking for a solution, but instead for a proof-check. If it's wrong I'll go back to the drawing board.
I tried using induction: assume that every $\beta<\alpha$ admits a subset of $(\mathbb{R},<)$ with order type $\beta$. Since $\alpha$ is assumed countable $\{\beta\in \text{Ord}:\beta<\alpha\}$ is countable, so we can enumerate as $\beta_0,\beta_1\dots$. Then map $f_k:\beta_k\to [k,k+1)\subset \mathbb{R}$ where $f_k$ order preserving and continuous, and $\sup f( \beta_k) = k+1$. Then $\bigcup \beta_k$ is a well-order with order type $\alpha$.
Does this work? If so, is it ok to claim the existence of such $f_k$'s?
The proof is incorrect. I think you meant to say that $\cup_kS_k$ has order-type $\alpha$ where $S_k$ is the image of $\beta_k$ under $f_k.$ But it's still wrong. For example if $\alpha=\omega + \omega=\omega \cdot 2$ then for some $k',k'',k'''$ we have $\beta_k'=\omega,\; \beta_k''=\omega +1, \beta_k'''=\omega + 2.$ Then each of the 3 intervals $[k',k'+1),\;[k'',k''+1),\;[k''',k'''+1)$ contains a subset of $\cup_kS_k$ that is order-isomorphic to $\omega,$ so the order-type of $\cup_kS_k$ is at least $\omega \cdot 3.$
For real numbers $a,b$ with $a<b$ let $g_{a,b}:[0,\infty) \to [a,b)$ be an order-isomorphic bijection.
For countable ordinal $\alpha ,$ suppose that for all $b<\alpha$ there exists an order-embedding $f_b$ of $b$ into $[0,1] .$ We show there exists an order-embedding of $\alpha$ into $[0,1].$
(i). If $\alpha =0 $ then $\alpha = \emptyset.$ Let $f_0=\emptyset.$ (Trivial special case.)
(ii). If $\alpha =c+1,$ let $f_c:c\to [0,1]$ be an order-embedding of $c$ into $[0,1] .$ Define $f_{\alpha} (x)=g_{0,1}(f_c(x))$ for $x <c,$ and $f_{\alpha }(c)=1.$ Then $f_{\alpha }$ is an order-embedding of $\alpha$ into $[0,1].$
(iii). If $0\ne \alpha =\cup \alpha,$ take $\{a_n: n\in \omega\}\subset \alpha$ with $a_n<a_{n+1}<\alpha$ for each $n\in \omega,$ and $\cup S=\alpha ,$ and $a_0=0.$
For each $n\in \omega$ the set $a_{n+1}$ \ $a_n$ is order-isomorphic to a (unique) ordinal $b_n,$ with $b_n<\alpha.$ So let $f_n$ be an order-embedding of $a_{n+1}$ \ $a_n$ into $[0,1].$
We have $\cup_{n\in \omega} (a_{n+1}$ \ $a_n)=\alpha$ (because $a_0=0$). For each $x\in \alpha$ there is a unique $n\in \omega$ such that $x\in a_{n+1}$ \ $a_n.$
$$\text {For } x\in a_{n+1} \backslash a_n \text { let }\quad h_{\alpha}(x)=g_{n,n+1}(f_n(x))$$ $$\text {and }\quad f_{\alpha}(x)=g_{0,1}(h_{\alpha}(x)).$$ Then $f_{\alpha}:\alpha\to [0,1]$ is an order-embedding.