Perhaps the most immediate argument to show that the closed disk is not homeomorphic to $\mathbb R^2$ is that the former is compact while the latter is not. This argument is not quite in line with the intuition you were trying to formalize. You intuition can be formalized to some degree by using some concepts from homotopy theory. The fundamental group of $\mathbb R^2$ with a point removed is $\mathbb Z$, while the fundamental group of the closed disk with a boundary point removed is trivial. Thus the two spaces are not even homotopy equivalent, let alone homeomorphic.
Thinking in terms of paths (basically the argument above but without assuming you know the fundamental group is), any path (that is a continuous function from $[0,1]$) in the disk with a boundary point removed can be continuously deformed into a point (just shrink the path to the centre of the disk). But, not every path in the plane with a point removed can be so contracted. If the path is a circle around the missing point then it can't be deformed to a point, the missing point keeps you from doing that.
The property of deformability of paths is an invariant of homeomorphisms (and of homotopy equivalences) and thus the two spaces are not homeomorphic.
Have a look at the catgorical definition of pushout. It is enough to show that $(V\times W)^\infty$ has the universal property. That's because pushouts are unique up to isomorphism.
I will use $V^\infty\vee W^\infty$, i.e. the wedge sum instead of $(V^\infty\times\{\infty\})\cup(\{\infty\}\times W^\infty)$ because thats what it is.
So we have a commutative diagram
$$\begin{CD}
V^{\infty}\vee W^{\infty} @>{i}>> V^{\infty}\times W^{\infty}\\
@VVV @VV{f}V\\
\{\infty\} @>{g}>> (V\times W)^\infty
\end{CD} $$
So $i$ is the inclusion, $g(\infty)=\infty$ and
$$f(v,w)=\left\{\begin{matrix}\infty &\text{if }v=\infty\text{ or }w=\infty \\ (v,w) &\text{otherwise}\end{matrix}\right.$$
So this is the first part of the universal property. I encourage you to check that the diagram commutes and that $f$ is continuous.
Lets have a look at the second part of the universal property. Assume that
$$\begin{CD}
V^{\infty}\vee W^{\infty} @>{i}>> V^{\infty}\times W^{\infty}\\
@VVV @VV{a}V\\
\{\infty\} @>{b}>> P
\end{CD} $$
is some commutative diagram. Define
$$\phi:(V\times W)^\infty\to P$$
$$\phi(v,w)=a(v,w)$$
$$\phi(\infty)=b(\infty)$$
So obviously $\phi\circ f=a$ and $\phi\circ g=b$. So all that is left is to show that $\phi$ is continuous. Can you complete the proof?
Side note: If you look carefuly at each function I've defined you will see that there's no magic here. These are just natural choices.
Best Answer
If $X$ is connected and $p \in X$ is such that $X\setminus \{p\}$ is disconnected, then $p$ is called a cut-point of $X$. Likewise, if $A \subseteq X$ is such that $X\setminus A$ is disconnected, $A$ is a cut-set of $X$. Easy fact: if $h: X \to Y$ is a homomorphism $A$ is a cut-set of $X$ iff $h[A]$ is a cut-set of $Y$.
So if $X \simeq Y$ they have the same numbers of cut-sets and (non-)cut-points. This is a rough but general tool to reason about certain connected spaces being homeomorphic or not:
So for your three spaces such simple considerations suffice.