Take $X,Y$ two pointed CW-complexes, with $X$ $k$-connected, $Y$ $l$-connected, we may also assume that $X$ or $Y$ is locally compact. I'm trying to show $(X\times Y, X\vee Y)$ is $(k+l+1)$-connected.
Here's where I'm at : I reckon that the hypothesis that one is locally compact is so that the product topology coincides with the weak topology. Furthermore the higher connectedness of $X$ implies that up to homotopy equivalence, $X$ can be assumed to have one $0$ cell, and no cells in dimensions $n$ for $0<n<k$. A similar statement can be made for $Y$. This gives a good handle on the low dimensions cells of $(X\times Y, X\vee Y)$. Namely, the product of the 0 cells of $X$ and $Y$ gives a $0$ cell of the product (right?). At first I thought we'd only have cells in dimension $k+l+1$ and higher but actually the cells $e_0\times e_\sigma$ where $e_0$ denotes the $0$ cell of either $X$ or $Y$ and $e_\sigma$ is either a $k+1$ cell of $X$ or an $l+1$ cell of $Y$, should give cells in the product of dimension lower than $k+l+1$. But I assume this has something to do with the fact that we're considering the relative CW complex $(X\times Y, X\vee Y)$. Regardless I'm not sure what more I can say, I'm not super familiar with the notions at play. Maybe this follows from Blackers-Massey theorem but I'm not sure.
Also, if $Z$ is a CW complex with no cells in dimension $k$ for $0<k<n+1$, can we say $Z$ is $n$ connected ? Intuitively I certainly feel like the answer is yes but I'm usually wrong more often than not.