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Any $n$-element set $X$ has exactly $2^n$ subsets $(n \ge 0)$?

Consider an arbitrary subset A of a given n-element set X, and define a mapping $f_{A}: X \to {0,1}.$ For an element $x \in X$ we put

$$f(n) = \begin{cases} 1 & \text{if $x \in A$} \\[2ex] 0 & \text{if $x \notin A$} \end{cases}$$

Distinct sets A have distinct mappings $f_{A}$, and conversely, any given mapping $f:X \to \{0,1\}$ determines a subset $A \subseteq X\text{ with} \ f = f_{A}$. Hence the number of subsets of X is the same as the number of all mappings $X \to \{0,1\}$, and this is $2^{n}$ by Proposition 3.1.1

I understand this proof generally, however, I still have not fully grasped the last part about the bijective correspondence. specifically its reference to the induction proof.

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    To construct all the subsets you start with the emptyset you pick each object one by one. Then you decide that you can either add it to each of the subsets you have so far or not. This doubles the number of subsets you have so far. Repeat with the next object. This is flimsy informal math. Counting is a matter of finding a 1-1 coresspondence between to sets. The argument here is that it is clearer to discuss the number of mappings. Each f:A->0,1 will correspond with precisely the a subset consisting of all the n so that f(n) = 1. So there are as many subsets as f's. – 2017-01-16
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    ... and for each f for each $a\in X$ wither f(a) = 1 or f(a) =0 there are $2^n$ such mappings. – 2017-01-16

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Let $F= \{f\}$ be the set of all functions that map $X\rightarrow \{0,1\}$.

Let $P = P(X) = \{$subsets of $X\}$.

Claim: There is a 1-1 correspondence between $F$ and $P$ and so they have the same cardinality.

Pf: If $f$ is a mapping $f:X \rightarrow \{0,1\}$ then we can define $A(f) = \{x \in X| f(x) = 1\} \subset X$. If $f \ne g$ then there is some $x \in X$ so that $f(x) \ne g(x)$. Without loss of generality we can assume $f(x) =1$ and $g(x) = 0$. So $x \in A(f)$ and $x \not \in A(g)$. So $A:F \rightarrow P(x)$ is injective. If $G \subset X$ then we can define a mapping $h: X \rightarrow \{0,1\}$ via $h(x) = 1$ if $x\in G$. $h(x) = 0$ if $x \not \in G$. So $G = A(h)$. So $A$ is surjective. So $A$ is bijection between $F$ and $P$.

Suffices to proof $|F|= 2^n$ and I presume that was the result of prop 3.1.1.