Prove that the power set of an $n$-element set contains $2^n$ elements

You misread the proof. Since set $X$ has $n$ elements, the induction hypothesis tells us that $|\mathcal{P}(X)| = 2^n$. The set $Y = X \cup \{a\}$ has $n + 1$ elements. It subsets are either subsets of $X$, of which there are $2^n$ by the induction hypothesis, or the union of a subset $Z$ of $X$ with $\{a\}$. By the induction hypothesis, there are $2^n$ subsets $Z$ of $X$. Hence, there are $2^n$ subsets of the form $Z \cup \{a\}$ of the set $Y$. Hence, $Y$ has $2^n$ subsets that do not contain $a$ and $2^n$ subsets that do contain $a$ for a total of $2^n + 2^n = 2 \cdot 2^n = 2^{n + 1}$ subsets of $Y$, which is what the author wants to show.

Here is an another, combinatorial proof. Let $X$ be a set with $n$ elements. To form a subset of $X$, we go over each element of $X$ and exercise a choice of whether or not to include in the subset. Every sequence of choices gives a different subset.

Since for every element, there are 2 choices, for $n$ elements, there are $2 \times 2 \times ...$ $n$ times, choices.

Therefore, there are $2^n$ distinct subsets of $X$.

Suppose you've already shown that $X=\{1,2\}$ has $2^2=4$ subsets, namely ${\cal P}(X)=\{\emptyset,\{1\},\{2\},X\}$. Now you add a new element $a=3$ to get $Y=X\cup \{a\}=\{1,2,3\}$. The four subsets of $X$ are also subsets of $Y$, but you get new subsets - those which contain $a$, e.g. $\{1,3\}$. But each of these is one of the ones you already had together with the new $a$, e.g. $\{1,3\}$ is $\{1\} \cup \{3\}$.

So, the new ones are $\emptyset \cup \{3\}$, $\{1\} \cup \{3\}$, $\{2\} \cup \{3\}$, and $X \cup \{3\}$ and those are exactly as much as you already had - four of them. Which implies that ${\cal P}(Y)$ has twice as much elements (the old ones and the new ones) as ${\cal P}(X)$, so $2\times2^2=2^3=8$.

$$ \begin{array}{|c|c|} \hline \text{Subsets of }X&\text{New subsets}\\ \hline \emptyset&\{3\}\\ \hline \{1\}&\{1,3\}\\ \hline \{2\}&\{2,3\}\\ \hline X&Y\\ \hline \end{array} $$