To answer first your questions. A) The statement is correct, and B) your proof is correct, including step $3)$.
I'm not sure of a good reference (I'd guess Hatcher has a thing to say about it, though), but a way to see what the attaching map is is as follows.
We know that $S^1\times S^1$ is a CW complex consisting of two 1-cells and a single 2-cell. Thus we have a cofibration sequence
$$S^1\xrightarrow{w}S^1\vee S^1\rightarrow S^1\times S^1$$
where $w$ is the attaching map and represents a class in $\pi_1(S^1\vee S^1)$. Using the Seifert-Van Kampen theorem we find that $\pi_1(S^1\vee S^1)\cong \mathbb{Z}\ast\mathbb{Z}$ is the free group on the two generators, $a=in_1:S^1\hookrightarrow S^1\vee S^1$, $b=in_2:S^1\hookrightarrow S^1\vee S^1$ given by the inclusions of the two respective wedge summands. Thus the homotopy class $w=a^{i_1}b^{j_1}\dots a^{i_r}b^{j_r}$ is some word in the generators $a,b$.
Now the abelianisation of this is just the free abelian group on these two generators $(\pi_1(S^1\vee S^1))_{ab}\cong\mathbb{Z}\oplus\mathbb{Z}$, and the Hurewicz theorem tells you that this is isomorphic to the homology $H_1(S^1\vee S^1)$. Now $H_1(S^1\times S^1)\cong\mathbb{Z}\oplus\mathbb{Z}$ also, and the inclusion of the 1-skeleton $S^1\vee S^1\hookrightarrow S^1\times S^1$ induces an isomorphism on homology. We see this using the homology exact sequence of the cofibration above
$$0\rightarrow H_2(S^1\times S^1)\xrightarrow{\cong} H_1(S^1)\xrightarrow{w_*} H_1(S^1\vee S^1)\xrightarrow{\cong} H_1(S^1\times S^1)\rightarrow 0.$$
The right hand map must be an isomorphism since both involved groups are free abelian and there is no torsion in the sequence. It follows that the left-hand map is also an isomorphism, since $H_2(S^1\times S^1)\cong\mathbb{Z}$ and $H_1(S^1)\cong\mathbb{Z}$ (they are oriented manifolds 2- and 1-manifolds respectively). Thus $w_*=0$ on homology.
The point of introducing the Hurewicz theorem earlier was that we may use its naturality to get a commutative diagram
$\require{AMScd}$
\begin{CD}
@>>>0@>>>\pi_1S^1@>\pi(w)>>\pi_1(S^1\vee S^1)@>>>\pi_1(S^1\times S^1)@>>>0\\
@V V V @VV V@VV \cong V @V V(-)_{ab}V@VV\cong V\\
0@>>> H_2(S^1\times S^1)@>>>H_1S^1@>w_*>>H_1(S^1\vee S^1)@>\cong>> H_1(S^1\times S^1)@>>>0
\end{CD}
Here the vertical maps are all Hurewicz morphisms. All three are abelianisations, and two turn out to be isomorphisms, using that $\pi_1(S^1\times S^1)=\pi_1S^1\oplus\pi_1S^1\cong\mathbb{Z}\oplus\mathbb{Z}$. $\pi(w)$ is the homomorphism induced by $w$. The bottom row of this diagram is exact, although the top is not necessarily so.
Since $\pi_1S^1$ is free abelian the homomorphism $\pi(w)$ will be determined by what it does on the generator, which is the identity $id_{S^1}$. Note that
$$\pi(w)[id_{S^1}]=[w\circ id_{S^1}]=[w].$$
Now we observed above that $w_*=0$, so the commutative diagram above tells us that $(\pi(w))_{ab}=0$. Thus $w$ lies in the kernel of the abelianisation homomorphism $ab:\mathbb{Z}\ast\mathbb{Z}\rightarrow \mathbb{Z}\oplus\mathbb{Z}$, which sends the free generators $a,b$ to the free abelian generators of the same name. The kernel of this map is the subgroup generated by commutators $[x,y]=xyx^{-1}y^{-1}$ in the words.
The point is that while $w$ is non-trivial in the non-abelian group $\pi_1(S^1\vee S^1)$, after suspending it becomes trivial in the abelian group $\pi_2(\Sigma (S^1\vee S^1))$ (recall that $\pi_2$ is always abelian).
In fact if $X$ is a path connected space then the Hurewicz map $h_1:\pi_1(X)\rightarrow H_1X$ is an abelianisation. However, after suspension $\pi_1\Sigma X=0$ so the Hurewicz map $h_2:\pi_2X\xrightarrow{\cong} H_1X$ is now an isomorphism. The suspension homomorphism $\Sigma:\pi_1X\rightarrow\pi_2\Sigma X$ actually identifies with the abelianisation, and this follows from naturality $h_2\circ \Sigma=\sigma\circ h_1$, where $h_2$ is the isomorphism above, $\sigma:H_1X\xrightarrow{\cong} H_2\Sigma X$ is the homology suspension, which is also an isomorphism, and $h_1$ is abelianisation.
Hence $\Sigma w=0\in\pi_2(\Sigma(S^1\vee S^1))=\pi_2(S^2\vee S^2)\cong\mathbb{Z}\oplus\mathbb{Z}$.
Incidentally, the map $w$ is a so-called Whitehead product, and takes the form
$$w=aba^{-1}b^{-1}.$$
You can see it has this form explicitly by identifying the torus $S^1\times S^1$ as a quotient of $I\times I$. The identification traces around the oriented boundaries, as indicated in Hatcher pg. 5.
More explicitly, Hatcher describes the product cell structure on a product of CW complexes on pg. 8. Then the cell structure on $S^1$ with one $0$-cell and one $1$-cell is given by identifying the two points in the boundary $S^0\cong \partial I$. Choosing a relative homeomorphism $(D^2,S^1)\cong (I\times I,I\times\partial I\cup \partial I\times I)$ and keeping track of the orientations it is clear that $w$ indeed has the form described.
The Whitehead product has higher dimensional generalisations that give attaching maps for the top cells of all products $S^n\times S^m$, and indeed more generally for any product of the form $\Sigma X\times \Sigma Y$.
The topology of the wedge sum is by definition the quotient topology.
The theory can be simplified a little bit: Let $(X,x_0)$ and $(Y,y_0)$ be pointed topological spaces, $X\sqcup Y$ the disjoint union, and $X\lor Y$ the wedge sum (with basepoints $x_0$ and $y_0$). Let $\pi\colon X\sqcup Y\to X\lor Y$ be the quotient map. Then a subset $A\subseteq X\lor Y$ is open if and only if $\pi^{-1}(A)$ is open. This is equivalent to saying that $\left\{x\in X:\pi(x)\in A\right\}$ ad $\left\{y\in Y:\pi(y)\in A\right\}$ are open in $X$ and $Y$, respectively. (I'm identifying elements of $X$ and $Y$ with their images in $X\sqcup Y$).
Since we are dealing with a wedge sum of $\mathbb{R}$ with itself, we can simplify the notation a bit: Given $x\in\mathbb{R}$, let $x_1$ be the class of $(x,0)$ in $\mathbb{R}\lor\mathbb{R}$ and $x_2$ be the class of $(0,x)$ in $\mathbb{R}\lor\mathbb{R}$. In this manner, $x_i$ is just the representative of $x$ in the $i$-th copy of $\mathbb{R}$. In particular, we have $0_1=0_2$.
Let $\pi\colon\mathbb{R}\sqcup\mathbb{R}\to\mathbb{R}\lor\mathbb{R}$ be the quotient map. Now let $A\subseteq\mathbb{R}\lor\mathbb{R}$ be open. This means that $\pi^{-1}(A)$ is open, so both sets $A_i=\left\{x\in\mathbb{R}:x_i\in A\right\}$ ($i=1,2$) are open in $\mathbb{R}$. The definition of $f$ yields $f^{-1}(A)=A_1\times\left\{0\right\}\cup\left\{0\right\}\times A_2$, which you can show to be open in $D$.
For example, if we have $x\in A_1\setminus\left\{0\right\}$, then $A_1\setminus\left\{0\right\}$ is open in $\mathbb{R}$, so $(A_1\setminus\left\{0\right\})\times\mathbb{R}$ is open in $\mathbb{R}^2$ and thus
$$((A_1\setminus\left\{0\right\})\times\mathbb{R})\cap D=(A_1\setminus\left\{0\right\})\times\left\{0\right\}$$
is open in $D$, contains $(x,0)$ and is contained in $f^{-1}(A)$. This proves that every nonzero point of $A_1\times\left\{0\right\}$ is interior to $f^{-1}(A)$. Similarly, every nonzero point of $\left\{0\right\}\times A_2$ is interior to $f^{-1}(A)$.
The only problem is to prove that $(0,0)$, if it belongs to $f^{-1}(A)$, is interior to $f^{-1}(A)$. But notice that $f(0,0)=0_1=0_2$, so $(0,0)\in f^{-1}(A)$ if and only if $0\in A_1$ and $0\in A_2$. In this case, $(A_1\times A_2)\cap D$ is an open neighbourhood of $(0,0)$ in $D$, which is therefore interior.
This shows that $f$ is continuous. To prove that $f$ is a homeomorphism, we can find its inverse: Let $g\colon\mathbb{R}\sqcup\mathbb{R}\to D$ be given by $g(x,1)=(x,0)$ and $g(x,2)=(0,x)$. Then $g(0,1)=(0,0)=g(0,2)$, so $g$ factors uniquely through a map $G\colon\mathbb{R}\lor\mathbb{R}\to D$, that is, to a map $G$ such that $g=G\circ\pi$. Since $g$ is continuous, then $G$ is also continuous (this is the universal property of quotients: If we have a quotient map $q\colon X\to X/\sim$, then a function $G\colon X/\sim\to Y$ is continuous if and only if $G\circ q$ is continuous).
It is easy to see that $G=f^{-1}$, so $f^{-1}$ is continuous.
Best Answer
Unfortunately, your proof can't be right since $\Sigma \mathbb{C}P^2$ is not homotopy equivalent to $S^3 \vee S^5$. This can be seen since the Steenrod square $\mathrm{Sq}^2$ is nontrivial on $H^3(\Sigma \mathbb{C}P^2; \mathbb{Z}/2)$. This is because there are in $H^2(\mathbb{C}P^2; \mathbb{Z}/2)$ with nontrivial squares.
The error in your proof is that you don't understand what attaching maps or CW structures on suspensions are. Particularly,
does not make sense since (a suspension of) the Hopf map should decrease degree by 1. As well,
is not a correct CW decomposition. If one is taking an unreduced suspension, the $S^1$ should be an interval. If one is taking a reduced suspension, it should just not appear.