Why is this "iterated compactification" construction on $\Bbb{R}^4$ so different from $S^4$? So the standard method of compactifying $\mathbb{R}^4$ is to append a point at infinity and then map $\mathbb{R}^4$ to $S^4$ using stereographic projection with the point at infinity representing the "north-pole" used in the stereographic construction.
The construction motivating this question is as follows:
Start by performing the standard compactification to $S^2$ on the copy of $\mathbb{R}^2$ comprising, say, the $xy$-plane. This transforms $\mathbb{R}^4$ so as to be a (trivial) plane bundle over $S^2\backslash \{\infty\}$ with a single point, $\infty$, which does not uphold a plane. Denote this sphere $S_0$ and its point at infinity by $\infty_0$. For each plane attached at some $p \in S_0\backslash\{\infty_0\}$, append a new point at infinity $\infty_p$ and perform the compactification process again. We now have uncountably many spheres $S_p$, one for each $p \in S_0 \backslash\{\infty_0\}$. Now "roll" each $S_p$ along $S_0$ so that in the end $\infty_p$ is identified with $\infty_0$; we can do this because there is a path $\gamma_0$ in $S_0$ from $p$ to $\infty_0$ and another path $\gamma_p$ in $S_p$ from $p$ to $\infty_p$, so "roll" $S_p$ along $S_0$ such that at time $t$ we have $\gamma_0(t)\sim \gamma_p(t)$. This realizes our space as something that seems a lot like an uncountable wedge of spheres.
̶I̶ ̶t̶h̶i̶n̶k̶ ̶t̶h̶e̶ ̶r̶e̶s̶u̶l̶t̶i̶n̶g̶ ̶s̶p̶a̶c̶e̶ ̶s̶h̶o̶u̶l̶d̶ ̶b̶e̶ ̶c̶o̶m̶p̶a̶c̶t̶,̶ ̶b̶u̶t̶ ̶I̶ ̶c̶o̶u̶l̶d̶ ̶b̶e̶ ̶w̶r̶o̶n̶g̶.̶
(This was silly, it is not.)
These constructions can't be homotopic since $\pi_2(S^4)=0$ whereas the uncountable wedge of spheres would need to have uncountable $\pi_2$, but this clearly isn't a regular ol' wedge of spheres (I underestimated how weird the topology would be).
I am wondering if there is some intuition for why this space is so different from $S^4$ when their underlying constructions seem very similar.
 A: Call the uncountably infinite wedge of spheres $X$ and let $\Phi: \Bbb{R}^4 \rightarrow X$ be the map realizing the construction in the problem statement.
Per Paul Sinclair's comments, $X$ is second countable. I was not initially convinced of this because it seemed we couldn't have a base yielding a neighborhood $N_p$ of $\infty_p$ satisfying $N_p \cap S_q = \emptyset$ for all $p\neq q$, but such sets don't correspond to any open set in our plane bundle and so cannot be members of the quotient topology. Rather, the open neighborhoods of $\infty_p$ will be uncountable unions of "would-be" open disks on the spheres (i.e. disks that would be open if we thought about equipping each sphere with the standard topology and then looked at the topology of their uncountable wedge), so specifying a neighborhood of any $\infty_p$ is the same as specifying a closed set whose intersection with the affine plane corresponding to $S_p$ is bounded. It is clear that $X$ is Hausdorff.
Let $P_0$ denote the cartesian plane, $\kappa_0: P_0 \rightarrow S_0$ the compactification map. Moreover, given $\kappa_0(x_p,y_p):=p\in S_0$, let $P_{p}$ be the plane attached to $\kappa_0(p)$ and let $\kappa_p : P_p \rightarrow S_p$ be its compactification. To show $X$ is not compact, consider the net $(N,\leq_{pre})$ defined by $$N:=\{z_{\alpha}\in S_{p_{\alpha}} \; | \; z_{\alpha} =(\kappa_{p_{\alpha}}\circ \kappa_0)(x_{p_{\alpha}},y_{p_{\alpha}},0,0), p_{\alpha} \in P_0\}$$ and preorder $$z_{\alpha}\leq_{pre} z_{\beta} 
 \Leftrightarrow x_{p_{\alpha}}^2+y_{p_{\alpha}}^2 \leq x_{p_{\beta}}^2+y_{p_{\beta}}^2$$
Such a net cannot accumulate at a point in $Im(\Phi)$ since this would imply there is a $q\in \Bbb{R}^4$ such that eventually (with respect to the preorder $\leq_{pre}$) all $(\kappa_{\alpha}\circ \kappa_0)^{-1}(z_{\alpha})$ are in $B(1,q)$ whereas I can select large enough $x$ and $y$ such that the distance between $(x,y,0,0)$ and $q$ exceeds $\epsilon$.
$N$ also cannot accumulate at the wedge point. Letting $A:=P_0\times [-1,1]^2\subset \Bbb{R}^4$, note that $\overline{\Phi(A)}^c$ is a neighborhood of wedge point and disjoint from $N$.
Hence, the net has no accumulation point and $X$ is not compact.
I am still curious if there is an interesting reason why these spaces end up being so different when the procedures for constructing them seem similar.
