Hello, I have been thinking of proving CRT since a while. It makes sense to me intuitively on why it is true as essentially \(\Bbb{Z}/{mn\Bbb{Z}}\cong \Bbb{Z}/{m\Bbb{Z}}\times \Bbb{Z}/{n\Bbb{Z}}, \gcd(m,n)=1\) so there is some isomorphism. However, let us prove it for ideals. Also, note that we can generalise the result to any number of ideals.

Let $R$ be a ring and $I$ and $J$ be ideals in $R$ such that $I+J =R$. So $I,J$ are co-maximal ideals.

Show that for any $r$ and $s$ in $R$, the system of equations \begin{align*} x & \equiv r \pmod{I} \\ x & \equiv s \pmod{J} \end{align*} has a solution.
We have $I+J=R$, so there exist $i \in I,j \in J$ such that $i+j=s-r$. Put $p=r+i=s-j$, and we get $p$ is a solution of the system since $p\in r+I$ and $p\in s+J$.
Prove that any two solutions of the system are congruent modulo $I \cap J$.
Say we get two different solutions $x$ and $x'$. Then \begin{align*} x &\equiv x' \pmod{I} \\ x &\equiv x' \pmod{J} \, , \end{align*} So $x - x'$ is in $I$ *and* $J$ $$\implies x - x' \in I \cap J\implies x \equiv x' \pmod{I \cap J}.$$
Let $I$ and $J$ be ideals in a ring $R$ such that $I + J = R$. Show that there exists a ring isomorphism $$ R/(I \cap J) \cong R/I \times R/J. $$
Note that by first part, we get the surjectivity. Using the isomorphism theorem and considering the map, \begin{align*} \phi: R &\rightarrow R/I \times R/J \\ x &\mapsto (x + I, x + J) \, . \end{align*} Note that The kernel of $\phi$ is when $x \in I,J$. We get that the kernel of the map is $I\cap J$. So $$R/(I \cap J) \cong R/I \times R/J.$$