Set of functions from $SS = \{ A_{1}, A_{2}, A_{3},...\}$ to $\{ \mathbf{T}, \mathbf{F} \} $ is countable? Let $SS=\{ A_1,A_2,A_3,\ldots\}$, and let $V = \{ v \mid v: SS \to \{ \mathbf{T}, \mathbf{F} \} \}$ .Is the set  V countable?  Justify your answer.
My instinct is to say that $V = \{ v \mid v: SS \to \{ \mathbf{T}, \mathbf{F} \} \}$ is uncountable, and to prove this using a diagonalization argument, i.e. create a table of values of the functions $v_{1}, v_{2},...$ for natural numbers $n_1, n_2,\ldots$ and define a function $v_{m} : SS \to \{ \mathbf{T}, \mathbf{F} \}$ where it takes all values in the diagonals of the table $T, F$ and flips them, and then show that $v_{m}$ cannot appear anywhere in the list. Is my guess correct, and if so would this be a reasonable approach to the problem?
Your approach is correct. Note that $V$ is bijective to the power set of $SS$ by mapping $v \in V$ to $v^{-1}(\mathbf{T}) \in 2^{SS}$, the subset of $SS$ that maps to $\mathbf{T}$.
Yes, this is (essentially) the countably infinite case of Cantor's Theorem.