Attempted proof of an open mapping theorem for Lie groups The Classical open mapping theorem for Banach spaces tells that if $T:X \to Y$ is a continuous surjective linear map, then it is open. 
I have attempted to essentially "adapt" the proof for Lie groups:

Let $G,H$ be connected Lie groups (embedded in $\mathbb R^n$, I've used second-countability and completeness so far). If $\phi: G \to H$ is
  a surjective Lie group homomorphism, then the image of an open
  neighborhood $U$ of the identity in $G$ is again a neighborhood with nonempty  interior
  about $1 \in H$.
Pick some neighborhood $U$ of $1 \in G$. We want to pick some open ball $V$in
  $U$ so that the closure of $V$ is compact and contained in $U$. Hence,
$$G=\{x V \mid x \in G\}$$
but we can choose countably many such $x \in G$.  We can call this
  collection $\{x_n\}$, and consider the image of $G$ under $\phi$,
  which is all of $H$. In particular, $H:=\{\phi(x_n \overline{V}) \mid n \in \mathbb N\}$. But since $\phi$ is a homomorphism, this is the
  same thing as considering a whole bunch of
  $\phi(x_n)\phi(\overline{V})$, whose images are compact and hence
  closed. By Baire category, one of these guys has empty interior, say
  $\phi(x_n)\phi(\overline{V})$. But then $\phi(\overline{V})$ has
  nonempty interior, since multiplication by an element is a
  homeomorphism. But then $\phi(U)$ has nonempty interior.

From here, one can finish by first showing that this implies that there is a basis $\mathcal U$ about the origin so that the image of each element contains the identity in $H$, which is sufficient to show the main theorem, since we can then use the homomorphism property to show that for every open set $W$ about $w \in G$, the image $f(w) \in \mathrm{Int}(W)$, proving the theorem.


*

*Is My proof correct?

*I've seen one reference "An Open Mapping Theorem for Topological Groups". but this ultimately just redirects to a source that I cannot find. Is there a better way to show this Theorem? I'm not looking for maximal generality (weakest hypotheses) just yet, only to understand why this theorem might be true.
A sufficient answer in my eyes is either a convincing argument for why my proof fails (including something in the way of "is this idea recoverable?") or some affirmation that it is indeed correct.
 A: When you have an action of a topological group $K$ over a space $X$, the quotient $X \to X/K$ is an open map. This is very easy to prove.
Now let $K$ be the kernel of your surjective map $\phi \colon G \to H$. The group $K$ acts on $G$ by multiplication on the right and the quotient is $G/K$. The map $\phi$ factors through a group homomorphism $ \bar{\phi} \colon G/K \to H$ which is bijective and continuous. If this map is open (hence an isomorphism of topological groups), then $\phi$ is open. So another way of looking at your question is, does the first isomorphism theorem hold for connected Lie groups? 
The answer is yes and one proof goes along the following lines:
1) The quotient $ G \to G/K$ is a submersion.
2) If $\phi$ is smooth, so is $\bar{\phi}$.
3) An injective homomorphism of Lie groups must be an immersion.
4) The map $\bar{\phi}$ is a submersion and an immersion, so
it is a local diffeomorphism, hence a diffeomorphism and so
an isomorphism of topological groups.
This is essentially the proof of Theorem 17.4 in 
https://www.staff.science.uu.nl/~ban00101/lie2016/lie2010.pdf
I could not see any flaw in your proof. I am not sure if you find that this way is better, but I found more intuitive to think about it in terms of the first isomorphism theorem (which does not hold in general for general topological groups).
