1- A balloon is exactly the same thing as $S^2$, so it is a manifold without boundary.
Your argument does not work: why would a neighbourhood of $P$ become contractible after removing $P$ ?
2- Your justification is correct, although you only really need the second part (indeed, half disks and disks remain connected when you remove a point)
3- It is not a manifold with boundary: the points that are identified have no neighbourhood that is a disk or a half disk (for the same reason as above)
Firstly, most of what you are asking can be found in Hewitt's and Ross' Abstract Harmonic Analysis: Volume I, Chapter 2, (8.13).
Secondly, I think @HennoBrandsma is correct, since for a topological group $T_0$ is not only equivalent to $T_2$ (being Hausdorff), but also to regularity [2, Chapter 2, 4.8] and even complete regularity [2, Chapter 2, 8.4]. Nevertheless, it is completely fine to assume in your definition that a topological group is Hausdorff, but it is precisely the regularity condition we need in the proof, since [1, p. 163, 2.3] gives us alternative characterizations of paracompactness for regular spaces (a similar statement can be found in Munkres' book, Lemma 41.3).
We may now proceed as in [2, Chapter 2, 8.13]:
Take a symmetric neighborhood $U$ of the identity element in $G$ such that $\overline{U}$ is compact (we used local compactness here). By [2, Chapter 2, 5.7]
$$L = \bigcup_{n = 1}^\infty U^n $$
is an open and closed group (notice that if $G$ is connected, $G = L$). Once we verify that $\overline{U} \subset U^2$, we get that
$$ L = \bigcup_{n = 1}^\infty \overline{U}^n $$
is a countable union of compact spaces, so $\sigma$-compact, and hence enjoys the Lindelöf property. Since left translations are homeomorphisms, every coset $xL$ is also Lindelöf.
Now take an arbitrary open cover $\mathcal{V}$ for $G$ and let $x \in G$. Obviously $\mathcal{V}$ also covers the coset $xL$, hence there exists a finite subcover $\{V_{xL}^{(n)}\}_{n = 1}^\infty$ of $\mathcal{V}$ for $xL$.
For every $n \in \mathbb{N}$ we define
$$ \mathcal{W}_n = \{ V_{xL}^{(n)} \cap xL ; \; xL \in G/L \} $$
and we claim that
$$ \mathcal{W} = \bigcup_{n = 1}^\infty \mathcal{W}_n $$
is a refinement of $\mathcal{V}$.
Indeed, clearly for every $x \in G$ and $n \in \mathbb{N}$ we have
$$ V_{xL}^{(n)} \cap xL \subset V_{xL}^{(n)} $$
Since every $x \in G$ is contained in precisely one coset $xL \subset G$, the family $\mathcal{W}_n$ is locally finite, hence $\mathcal{W}$ is a $\sigma$-locally finite refinement of $\mathcal{V}$ that itself covers $G$.
By the above alternative characterizations of paracompactness for regular spaces, $G$ is paracompact.
Note: Since every paracompact Hausdorff space is normal, this effectively shows that every locally compact (Hausdorff) topological group is normal.
Sources:
[1] J. Dugundji, Topology, 12th printing ed., Allyn Bacon, 1978.
[2] E. Hewitt and K. A. Ross, Abstract Harmonic Analysis I, Springer-Verlag, 1979.
Best Answer
Let $\sim$ be the binary operation corresponding to path-connectivity in $SL_n(\mathbb{R})$; by my answer here, $\sim$ is an equivalence relation.
In order to show $SL_n(\mathbb{R})$ is path-connected, it suffices to show $A\sim I_n$ for all $A\in SL_n(\mathbb{R})$. But by my answer here, $A$ can be written as a (possibly empty) product of elementary matrices of the first type, so it in fact suffices to prove that$$E_{uv}(a)M\sim M$$for all $M\in SL_n(\mathbb{R})$ and Type 1 elementary matrices $E_{uv}(a)$ ($1\le u,\,v\le n$) of the form$$I_n + [a[(i,\,j) = (u,\,v)]]_{i,\,j\,=\,1}^n.$$Yet$$M\to E_{uv}(b)M$$simply adds $b$ times row $j$ to row $i$, i.e. takes $r_i$ to $r_i+br_j$. For fixed $u$, $v$, $M$, this map is continuous in $b$ (and preserves the determinant), so the continuous function$$X(t) = E_{uv}(ta)M$$over $[0,1]$ takes$$X(0) = M \to X(1) = E_{uv}(a)M$$while remaining inside $SL_n(\mathbb{R})$, as desired.