In Example 1.1 of https://maths.dur.ac.uk/users/andrew.lobb/master_morse.pdf it explains why $\mathbb{C}P^2$ and $S^2\vee S^4$ are not stably homotopy equivalent by looking at the second Steenrod squares.
I understand "stable homotopy equivalent" as in https://etale.site/livetex/wcatss/sht1.pdf. That is, $X$ and $Y$ are stable homotopy equivalent if there is a large $m<\infty$ such that the $m$-th suspension of both are homotopy equivalent. In particular, note $S^1$ and $S^2$ would not be stably homotopy equivalent in this sense.
I also understand that we can see the second Steenrod squares of both spaces are different because in second cohomology the operation coincides with cup product squaring; if $\alpha$ is the generator of second cohomology, in $\mathbb{C}P^2$ $Sq^2(\alpha)$ equals the top cohomology generator, and in $S^2\vee S^4$ it equals zero.
We also know the second Steenrod square is a stable cohomological operation, that is, it commutes with suspensions, as in https://amathew.wordpress.com/2011/11/16/the-steenrod-algebra-and-its-dual/#more-2962. I understand that this would mean $\Sigma^m (Sq^2_X(\alpha)) = Sq^2_{\Sigma^m X}(\Sigma^m \alpha)\in H^{(m+2)+2}(\Sigma^m X)$, where $\alpha\in H^2(X)$.
Question: How do we use this information (the two spaces $\mathbb{C}P^2$ and $S^2\vee S^4$ have distinct second steenrod squares) to show they are not stable homotopy equivalent?
In other words, how do we show from the bare fact that Steenrod squares commute with suspensions that a Steenrod square is a stable homotopy invariant? Of course, saying the Steenrod squares are stable cohomology operations is begging the question in this context (i.e. my question is about chasing definitions).
Side Question: I feel like some of my "I understand" statements above are not accurate. Please let me know if my definitions are not right.
Best Answer
If we know that the action of $\mathop{Sq}^2$ is a homotopy invariant and that it is also stable (compatible with suspension), then we argue by contradiction as follows: assume that there is an $m$ so that $\Sigma^m \mathbb{C}P^2$ and $\Sigma^m (S^2 \vee S^4)$ are homotopy equivalent. We of course know their mod 2 cohomology as vector spaces, and the fact that $\mathop{Sq}^2$ is a homotopy invariant means that we could conclude that this diagram commutes (all cohomology with mod 2 coefficients): $$ \begin{array}{ccc} H^{m+2} \Sigma^m \mathbb{C}P^2 & \xrightarrow{\cong} & H^{m+2} \Sigma^m (S^2 \vee S^4) \\ \downarrow^{\mathop{Sq}^2} & & \downarrow^{\mathop{Sq}^2} \\ H^{m+4} \Sigma^m \mathbb{C}P^2 & \xrightarrow{\cong} & H^{m+4} \Sigma^m (S^2 \vee S^4) \end{array} $$ The suspension isomorphism and the stability of $\mathop{Sq}^2$ then gives us commuting squares on the left and right sides of this one: $$ \begin{array}{ccccccc} H^2 \mathbb{C}P^2 & \xrightarrow{\cong} & H^{m+2} \Sigma^m \mathbb{C}P^2 & \xrightarrow{\cong} & H^{m+2} \Sigma^m (S^2 \vee S^4) & \xrightarrow{\cong} & H^2 (S^2 \vee S^4) \\ \downarrow^{\mathop{Sq}^2} & & \downarrow^{\mathop{Sq}^2} & & \downarrow^{\mathop{Sq}^2} & & \downarrow^{\mathop{Sq}^2} & & \\ H^4 \mathbb{C}P^2 & \xrightarrow{\cong} & H^{m+4} \Sigma^m \mathbb{C}P^2 & \xrightarrow{\cong} & H^{m+4} \Sigma^m (S^2 \vee S^4) & \xrightarrow{\cong} & H^4 (S^2 \vee S^4) \end{array} $$ Since the resulting diagram does not actually commute (the square formed by the left and right sides of the whole diagram), then the initial assumption must be wrong: we cannot have an $m$ so that $\Sigma^m \mathbb{C}P^2$ and $\Sigma^m (S^2 \vee S^4)$ are homotopy equivalent.