Is it not clear that if you call $(r_n)$ the dyadic rational whose $k$-th binary digit is $1$ when $k≤n$ and the $k$-th Turing machine halts in less than $n$ steps, and $0$ otherwise, thus defining a Specker sequence, then $\{r_n|\}$ as a surreal (computable in your sense) equals $\sup(r_n)$ as a real, which is not computable? Your paragraph following question (2) suggests you saw this argument but have at least a little doubt about it: to me it seems convincing, no?
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