So given a system of simultaneous congruences we know by CRT there exists a unique solution modulo the product of all the individual moduli.
Here is my proof:
Let $N=n_1\cdot n_2\cdots n_k$ where all the $n_i$'s are pairwise coprime, and let $N_i=N/n_i$ for $1\leq i\leq k$. Since all the $n_i$'s are pairwise coprime, $N_i$ is also coprime to $n_i$. So by definition there exists an inverse $x_i$ modulo $n_i$, that is we have $x_iN_i\equiv1\pmod{n_i}$. We also have $x_iN_i\equiv0\pmod{n_j}$ for $i\not= j$. Consider the following $x=x_1N_1a_1+x_2N_2a_2+\cdots+x_kN_ka_k$. Consider this equation modulo $n_i$. Then every term where the subscript is not $i$ will be $0$ from the implications we made and every other term where the subscript is $i$ we know the $x_i$'s and $N_i$'s will be $1$. We have $x\equiv 0 +\cdots 0 + x_iN_ia_i+0+\cdots +0\pmod{n_i}$. But since we know $x_iN_i\equiv 1\pmod{n_i}$, we have constructed our solution, namely $x\equiv a_i\pmod{n_i}$ for $1\leq i\leq k$.
This shows existence of a solution, now for uniquness,
Suppose $x$ and $y$ are solutions such that $x\equiv a_i\pmod{n_i}$, $y=a_i\pmod{n_i}$, for all $i$ in $1\leq i\leq k$. This tells us $x-y\equiv 0\pmod{n_i}$ for all $i$ in $1\leq i\leq k$. Then by definition we have $n_i|(x-y)$ for all $i$ in $1\leq i\leq k$. But recall the $n_i$ are pairwise coprime, hence by unique factorization we have $n_1n_2\dots n_k|(x-y)$, that is $N|(x-y)$ and by definition we have $x\equiv y\pmod{N}$. This proves uniqueness of the solution we showed in 2.1.
Question #1: Is my proof constructed well, is there any issues?
Question #2: I am relatively new to number theory, but I thought the solution is a solution where the modulo is a product of all the $n_i$'s, but this proof suggests the modulo of the solution is just a single $n_i$?