Is methanol really more acidic than water?
First, water's $\ce{pK_a}$ in water is $\ce{14}$ as explained here. This means that $\ce{H2O}$ is slightly dissociated in liquid form, such that $\ce{[H+] = [HO-] = 10^{-7}}$.

Second, I would say that caring too much about which compound is the best acid is a bit like arguing by definition. Acidity is a fuzzy concept used as a shorthand for deeper meaning. This means there is little point arguing about the shorthand when we have the actual properties it synthesizes.

In this case (sou rce):

$$
    \begin{array} {|c|c|c|l|}
    \hline
    \ce{solvent} &amp; \ce{pK_a\ of\ water} &amp; \ce{pK_a\ of\ methanol} &amp; \ce{Interpretation} \\
    \hline
    \ce{water} &amp; 14.0 &amp; 15.5 &amp; \ce{water\ is\ more\ dissociated\ than\ methanol} \\
    \hline
    \ce{DMSO} &amp; 31.4 &amp; 29.0 &amp; \ce{methanol\ is\ more\ dissociated\ than\ water}\\
    \hline
    \end{array}
$$
So you can argue (as in the answer you linked) that methanol is more acidic than water in the abstract sense, because $\ce{DMSO}$ doesn't stabilize the anions by hydrogen bonding and so is closer to the "ideal" case (whatever that may be).

But what's the point ? You already have the $\ce{pK_a}$ values, any practical question you might have about the concentrations of theses solutes in $\ce{H2O}$ and $\ce{DMSO}$ is already answerable.