The proposition is 1.11 from the commutative algebra book.
Let $\mathfrak{p}_1, \dots, \mathfrak{p}_n$ be prime ideals and let $\mathfrak{a}$ be an ideal contained in the union of those prime ideals. Then $\mathfrak{a} \subset \mathfrak{p}_i$ for some $i$.
This is proved by induction on $n$ of the form $\mathfrak{a} \not\subset \mathfrak{p}_i (i =1\dots n) \implies \mathfrak{a} \not\subset \bigcup \mathfrak{p}_i$. It is certainly true for $n = 1$. If $n \gt 1$ and the result is true for $n-1$, then for each $i$ there's $x_i \in \mathfrak{a}$ such that $x_i \notin \mathfrak{p}_j$ whenever $i \neq j$. [...]
I don't see the logic in this. Let $x_i \in \mathfrak{a}$ such that $x_i \notin \mathfrak{p}_i$, for all $i \neq j$, but what if $\mathfrak{p}_i \subset \mathfrak{p}_j$ for some $j$?
That's the induction hypothesis. The case $\mathfrak p_i\subset\mathfrak p_j$ poses no problems since there is no claim that $x_i$ belongs to $\mathfrak p_i$.