I don't know whether it would be of enough interest, but here is a relatively simple method for seeing why it must not be true. I posted almost the same method in a question last year (which I later deleted because there was an incorrect assumption in the question).

The method only uses natural numbers. The natural numbers are supposed to represent index of ordinal programs. Let $\lambda$ be the supremum of clocking positions (empty input). We are interested in all ordinals of the form $\beta_i$ such that: (i) $\beta_i$ is addmissible (or limit of admissibles) (ii) the "count" of clocking positions below $\beta_i$ is also equal to $\beta_i$.

Now we simply have a (ordinal) counter $s$ which starts from $0$ and keeps on increasing. At each value of the form $\beta_s$ we have the set $A_s \subseteq \mathbb{N}$ (with $A_0=\mathbb{N}$). Interpreting each $e \in A_s$ as an index of a program from $\beta_s$ to $\{0,1\}$, to form $A_{s+1}$, we remove all those indexes which represent total $\beta_s$-computable functions [total over the domain $\beta_s$] calculating linear-orders (that aren't well-orders). At limit values of $s$, we have $A_s$ formed as intersection of all $A_i$'s where $i<s$. Also, by construction we have $A_i \supseteq A_j$ whenever $i<j$, meaning any element that is removed once from the list is removed permanently.

The main thing to note is that for each value $\beta_i$ we have $sup(\beta_i \mbox{-} \mathrm{computable})$ equal to $sup(\beta_i \mbox{-} \mathrm{zeroComputable})$. Here by latter I mean the supremum of $\beta_i$-computable well-orders without any parameters. Finally, when we have $\beta_{n}=\lambda$ we must have $A_n=A_{n+1}$ [because otherwise we would have a program, on empty input, clocking above $\lambda$]. But now if $\lambda$ was a good ordinal then there would be a $\lambda^+$-computable function from $\alpha<\lambda^+$ [e.g. $\alpha=\lambda+\omega$] to $\lambda^+$ with co-finality $\lambda^+$. If we write the bad ordinals enumerated by this method above as $b_i$ then there one thing that we note is that $b_\omega \neq sup\{b_i|i < \omega\}$.

One slight generalization of this method would be to consider all programs (with parameters) as each suitable stage (ignoring the condition of clocking positions mentioned above). But then we wouldn't have the set $A_s$ as just containing natural numbers, but instead $A_s$ would keep growing as $s$ increases. One benefit of such a method would be that it would show any sufficiently closed ordinal to be bad. For example, on top of $\lambda$ the method would also show $\omega_1$ to be bad.