Is reaching in less lines semi-decidable? Assuming we have two programs $p_1$ and $p_2$ and two line numbers $n_1$ and $n_2$. Does $p_1$ reach $n_1$ in less computational steps than $p_2$ reaches $n_2$? By reduction from Halting, this is clearly not decidable, but I think it is semi decidable.

To do so, I would build an interpreter, that executes $p_1$ and $p_2$ simultaneously step by step and count the steps for each program. As soon as $p_1$ reaches $n_1$, I compare the number of steps to $n_2$ and return true if it is less. If $p_2$ reaches $n_2$ first, I return false. In case no program reaches $n_1$ or $n_2$, nothing happens (following semi-decidability).
It's only semi-decidable if you word your decision problem very carefully. And you have to word it in such a way that the case where both programs never reach their $n$s is in the REJECT category. Since $\infty &lt; \infty$ is ambiguous/undefined I would explicitly mention this case in your decision problem.

Other than that, yes it's correct. Your machine halts if either $p_1$ or $p_2$ halts (viewing reaching $n$ as just another halting condition) and provides the correct answer in such a scenario. If you patch up the above issue then when neither $p_1, p_2$ halts you are in the REJECT case and thus you are allowed to never halt as well by semi-decidability.