Does there exist a smooth structure on $[0,\infty)$ that induces the standard topology? I am sure that there is a problem with the point 0, but I don't know how to show it rigorously.
 A: No. Assume there exists a smooth structure on $M = [0,\infty)$ that induces the standard topology. Then $M$ would be an $n$-dimensional smooth manifold for some $n$. In particular $M$ would be locally Euclidean with "model space" $\mathbb R^n$.
Thus for each $p \in M$ there would exist a homeomorphism $h :  U \to B^n$ from an open neigborhood $U$ of $p$ (in the standard topology!) to the open unit ball $B^n$ in $\mathbb R^n$ such that $h(p) = 0$. Note that $U$ must be connected, i.e. an interval.
Let us now prove that we must have $n=1$. Let $x$ be an interior point of the interval. Then $U \setminus \{x\}$ is disconnected, thus also $h(U \setminus \{x\}) = B^n \setminus \{h(x)\}$ is disconnected. But for $n \ge 2$ this cannot happen.
Now take $p = 0$. Then $U$ must be an interval of the form $[0,a)$ with $0 < a \le \infty$. But $U \setminus \{0\}$ is connected, thus also  $h(U \setminus \{0\}) = B^1 \setminus \{h(0)\} =(-1,0) \cup (0,1)$ is connected, which is not true. This contradiction shows that the above assumption was false.
