Cardinality of a basis for $\prod _{\gamma<\alpha} [0,1]_\gamma$ Claim for any infinite ordinal $\alpha$ there exists a basis for $\prod _{\gamma<\alpha} [0,1]_\gamma$ of size $|\alpha|$.  
proof. Each factor $[0,1]$ has a basis of cardinality $\omega$ from which we can get a subbasis for $\prod _{\gamma<\alpha} [0,1]_\gamma$ of size $|\alpha\times\omega|=|\alpha|$, leading to a basis which also must have size $|\alpha|$.
Is my argument correct?  Also, does the product of $\alpha$-many spaces each with a basis of cardinality $\leq |\alpha|$ have a basis of size $\leq |\alpha|$?

Yes.
Let $\kappa$ be an infinite cardinal. Suppose that for each $\xi<\kappa$ the space $X_\xi$ has a base $\mathscr{B}_\xi$ of cardinality at most $\kappa$. Let $X=\prod_{\xi<\kappa}X_\xi$, and for $\xi<\kappa$ let $\pi_\xi:X\to X_\xi$ be the canonical projection map. Then
$$\left\{\pi_\xi^{-1}[B]:\xi<\kappa\text{ and }B\in\mathscr{B}_\xi\right\}$$
is a subbase for the topology of $X$ of cardinality $\kappa$, and its closure under finite intersections is a base of cardinality $\kappa$.
