Help understanding the argument used in this proof that $SO(n)$ is path-connected Background: I am reading Tapp's matrix groups for undergraduates and I am in the process of showing that $SO(3)$ is path-connected. While working on my own arguemnt I found a proof online in these lecture notes. 
But I need some help understanding the argument. 
The proof is this:
$SO(1) = \{1\}$ so for $n=1$ we see that $SO(n)$ is path-connected. Assume that $SO(n-1)$ is path-connected. 
Consider the continuous action of $SO(n)$ on $\mathbb R^n$ by left multiplication. The stabilizer of $e_n$ is $SO(n − 1) \le SO(n)$ thought of as the closed subgroup of matrices of the form
$$\begin{pmatrix}
P & 0\\
0^T & 1
\end{pmatrix}$$
with $P \in SO(n − 1)$ and $0$ the $(n − 1) \times 1$ zero matrix. The orbit of $e_n$ is the unit sphere $S^{n−1}$ which is path connected. Since the orbit space is also diffeomorphic to $SO(n)/ SO(n − 1)$ we have the inductive step.

I have some questions about this proof:
(1) What is $e_n$? The standard basis of $\mathbb R^n$?
(2) Is there a theorem "If a group $G$ acts continuously on $\mathbb R^n$ and the orbits of the basis elements are path-connected then so is $G$"?
(3) Where in the proof is it used that the stabilizer is $SO(n-1)$?
(4) I assume that since we showed that the orbit space is
  path-connected and we assume that it is diffeomorphic to
  $SO(n)/SO(n-1)$ then $SO(n)/SO(n-1)$ is also path-connected. Is there
  a theorem that "If $X$ is a topological space and $Y$ a path-connected
  subspace and the quotient space $X/Y$ is path-connected then $X$ is
  path-connected"?

 A: (1) $e_n$ is the last element of the standard basis. That is, $e_n = (0,0,0,\ldots,0,1)$.
(2/3) I've combined these together, because this is the theorem I think you want to ask: If $H$ and $G/H$ are connected, is $G$? The answer is yes, under some technical assumptions that work out here. (This is also where we need the fact that the stabilizer is $SO(n-1)$.) We need the quotient map $\pi:G \rightarrow G/H$ to have the path lifting property. So if $p,q \in G$, we connect $\pi p$ and $\pi q$ with a path in $G/H$ and lift that to $G$, giving a path from $p$ to $q'$, where $\pi q' = \pi q$. Since $q',q$ have the same orbit, they differ by an element of $H$, which is path connected. Hence we can connect $q'$ and $q$ by a path.
(4) Something you should be aware of, if you aren't already: If $H \subseteq G$ is a pair of topological groups with $H$ normal, then the group quotient $G/H$ gets a natural topology, and that is the topology used in the previous questions.
There is also a topological quotient that forgets everything about normal subgroups and orbits and such and simply says "Hey, here's a pair of spaces $H \subseteq G$ and I can squish $H$ to a point." This doesn't look like the group quotient above, since only things in $H$ are identified, rather than things in the same orbit.
The theorem you're asking about in (4) is asking about the second sorf of quotient, and has some counterexamples that probably all look sort of weird, like this one: Let $X$ be the arc of the unit circle with angle $0 \leq \theta \leq \pi$. Add to it line segments from the arc towards the origin starting at angle $\frac{\theta}{n}$ and having length $1-\frac{1}{n}$. Also include the origin. This space is not path connected, since there's no way to reach the origin from the rest of the space. However, $X\setminus\{(0,0)\}$ is a path connected subspace, and the quotient space is a two point space $\{o,y\}$ with open sets $\{y\}, \{o,y\}, \emptyset$. The function $$f(t) = \begin{cases} y \mbox{ if } t < \frac{1}{2} \\ o \mbox{ else } \end{cases} $$ is continuous, oddly enough... and so the quotient space is path connected.
