Can a set (as a candidate to be a vector space) be shown to be closed under vector addition and scalar multiplication in one step?

Yes.

The latter two statements are specialisations of the first statement; the first statement being true implies that they are true, if you set $(\alpha = \beta = 1)$ and then $(\beta = 0)$ while $(\alpha)$ is allowed to take on an arbitrary value.

The truth of the first statement is implied from the latter two statements with "$(\alpha \textbf{v})$ and $(\beta \textbf{w})$ are in $V$" as an intermediate logical step.