We define the half-space $H^n$ as the set containing all tupels $(a_1,\ldots,a_n)$ such that all $a_i\geq 0$.
I know that this isn't a manifold - intuitively this is clear - but how can I formally prove it ? I tried to prove the case $n=1$ via contradiction, but couldn't find a way to derive a contradiction, so I'm thinking that there maybe are some calculus facts about continuous maps that I don't know...
Hint: let $n = 2$ for concreteness, so we get the idea. As you wrote, $H^2$ will be the first quadrant of the cartesian plane. If this is to be a manifold, then it will be at most $2$-dimensional. If you take points $(a_1,a_2)$ with $a_i > 0$ nothing will go wrong, you have neighbourhoods in $H^2$ homeomorphic to disks in $\Bbb R^2$. To see that it is not a $2$-manifold, consider a point, say $(a_1,0)$, with $a_1 > 0$. Is there a neighbourhood of this point, in $H^2$, which is homeomorphic to an open disk in $\Bbb R^2$? Try to generalize.