The trace only depends on values near the boundary. That is, if $\phi\in C^\infty([0,\infty))$ is one in a neighborhood of zero, then $\operatorname{tr}_\Omega(\phi u)=\operatorname{tr}_\Omega(u)$ for every $u\in H^1(\mathcal C)$. With this in mind, you can formally apply a cut-off to your constant function and treat the trace in that way. Traces do indeed make sense in $H^\epsilon(\mathcal C)$ since multiplication by a compactly supported function brings your function to $H^1(\mathcal C)$.

But this is not needed if you just want to show that the inclusion is strict. The point is that $H^\epsilon(\mathcal C)$ is a completion of $H^1(\mathcal C)$. When showing that constant functions are in the completion but not the original space, we should of course remember that they are not in the original space – if you permit the tautology. The norm $\|\cdot\|_{H^\epsilon(\mathcal C)}$ was only defined for functions in $H^1(\mathcal C)$ by the integral expression in the first place, so the expression $\|u_n-1\|_{H^\epsilon(\mathcal C)}$ (or $\|1\|_{H^\epsilon(\mathcal C)}$) indeed does not make sense for the norm defined on $H^1(\mathcal C)$. (Of course the norm can be naturally extended and the integral expression is exactly the same.)

Once you have confirmed (as you seem to have) that your sequence is Cauchy in the norm, then it automatically has a limit in the completion. The sequence converges locally uniformly (in fact, it is eventually constant in any compact set), so it is easy to observe that if it had a limit in $H^1(\mathcal C)$, it would have to be the pointwise limit – the constant function. But the constant is not in $H^1(\mathcal C)$, so you have indeed shown that the completion contains a point outside the original space. This point is a point in a formal completion (an equivalence class of Cauchy sequences) but in this case it is natural to identify with a function (which is not in $H^1(\mathcal C)$).

The trace only depends on values near the boundary. That is, if $\phi\in C^\infty([0,\infty))$ is one in a neighborhood of zero, then $\operatorname{tr}_\Omega(\phi u)=\operatorname{tr}_\Omega(u)$ for every $u\in H^1(\mathcal C)$. With this in mind, you can formally apply a cut-off to your constant function and treat the trace in that way. Traces do indeed make sense in $H^\epsilon(\mathcal C)$ since multiplication by a compactly supported function brings your function to $H^1(\mathcal C)$.

But this is not needed if you just want to show that the inclusion is strict. The point is that $H^\epsilon(\mathcal C)$ is a completion of $H^1(\mathcal C)$. When showing that constant functions are in the completion but not the original space, we should of course remember that they are not in the original space – if you permit the tautology. The norm $\|\cdot\|_{H^\epsilon(\mathcal C)}$ was only defined for functions in $H^1(\mathcal C)$ by the integral expression in the first place, so the expression $\|u_n-1\|_{H^\epsilon(\mathcal C)}$ (or $\|1\|_{H^\epsilon(\mathcal C)}$) indeed does not make sense for the norm defined on $H^1(\mathcal C)$. (Of course the norm can be naturally extended and the integral expression is exactly the same. You just need to extend the trace map to $\operatorname{tr}_\Omega:H^\epsilon(\mathcal C)\to L^2(\Omega)$ via cut-offs or otherwise to make sense of the expression.)

Once you have confirmed (as you seem to have) that your sequence is Cauchy in the norm, then it automatically has a limit in the completion. The sequence converges locally uniformly (in fact, it is eventually constant in any compact set), so it is easy to observe that if it had a limit in $H^1(\mathcal C)$, it would have to be the pointwise limit – the constant function. But the constant is not in $H^1(\mathcal C)$, so you have indeed shown that the completion contains a point outside the original space. This point is a point in a formal completion (an equivalence class of Cauchy sequences) but in this case it is natural to identify with a function (which is not in $H^1(\mathcal C)$).