Proof
Enumerate S = {\(x_1, x_2, ..., x_{|S|}\)} and note \(P( \{ x_k \} ) = c\) for some constant c since each item is equally likely. However, using each outcome as a disjoint event and the definition of probability,
\begin{align*}
1 = P(S) & = P( \{ x_1 \} \cup \{x_2 \} \cup ... \cup \{x_{|S|} \} )\\
& = P(\{ x_1 \}) + P(\{ x_2 \} ) + ... + P(\{ x_{|S|} \} )\\
& = c + c + ... + c = {|S|} \times c
\end{align*}
and so \(c = \frac{1}{{|S|}}\text{.}\) Therefore, \(P( \{ x_k \} ) = \frac{1}{|S|}\) .
Hence, with A = {\(a_1, a_2, ..., a_{|A|}\)}, breaking up the disjoint probabilities as above gives
\begin{align*}
P(A) & = P( \{ a_1 \} \cup \{ a_2 \} \cup ... \cup \{ a_{|A|} \} )\\
& = P(\{ a_1 \}) + P(\{ a_2 \} ) + ... + P(\{ a_{|A|} \} )\\
& = \frac{1}{{|S|}} + \frac{1}{{|S|}} + ... + \frac{1}{{|S|}}\\
& = \frac{|A|}{{|S|}}
\end{align*}
as desired.