Let $R$ be a ring. If $I_1,\ldots, I_k$ are ideals of $R$ and for all $i$:
$$I_i + \bigcap_{j\neq i} I_j = R$$
then for all $a_1,\ldots,a_k \in R$, there exists some $a \in R$ such that for all $i$, $a \equiv a_i \pmod{I_i}$
I could prove this for $k=2$: for $a_1,a_2 \in R$, $a_1 - a_2 \in I_1 + I_2$, so there are $i_i \in I_i$ such that $a_1 - a_2 = -i_1 + i_2$. Hence $a_1 + i_1 = a_2 + i_2 = a$, and $a \equiv a_i \pmod{I_i}$ for $i=1,2$.
I couldn't prove the general case. Any help?
On
Hint $ $ It is easy if we view the solutions as "vectors" in the product ring $\, (R/I_1,\ldots, R/I_k).\,$ Then every such congruence is solvable iff the natural image of $R$ in the product ring is onto iff the image contains all unit vectors $e_j$ with $1$ in the $j$'th component and $0$ in all others, since they span, e.g. $$(a_1,a_2,a_3) = a_1(1,0,0)+a_2(0,1,0)+ a_3(0,0,1) = a_1 e_1 + a_2 e_2+ a_3 e_3$$
But the hypotheses imply that unit vector systems are solveable, since by hypothes $\,a+b = 1\,$ for $\,a\in I_j,\ b\in I_i, \forall\, i\neq j$ so $\,e_j := 1\!-\!a = b\,$ $\,\Rightarrow\,e_j\equiv 1\pmod{I_j},\ $ $e_j\equiv 0\pmod{I_i},\forall\,i\ne j$
Remark $ $ The classical CRT formula for integers is a special case, where the unit vectors are constructed by the formula
$\qquad e_j := M_j(M_j^{-1} \bmod m_j)\,$ where $M_j$ is the product of all moduli except $m_j$
Note $\,M_j \equiv 1\pmod{m_j}\,$ and $\,M_j\equiv 0\,$ mod other moduli since they divide $\,M_j$
Hint for the inductive step:
Suppose $I_1,\dots, I_k,I_{k+1}$ are ideals in $R$ such that for any $1\le i \le k+1$, $$I_i+\bigcap_{\substack{1\le j \le k+1\\j\ne i}}I_j=R. $$
Then, a fortiori, we have for any $1\le i \le k$, $$I_i+\bigcap_{\substack{1\le j \le k\\j\ne i}}I_j=R. $$ So, given $a_1, \dots, a_k, a_{k+1}\in R$, there exists some $a\in R$ such that $$a\equiv a_i\mod I_i\quad(1\le i\le k)$$ and $a$ is unique mod. $\displaystyle\bigcap_{j=1}^k I_j$.
Can you end the inductive step?
Some more details:
Note that if $J=\displaystyle\bigcap_{j=1}^k I_j $, we have $J+I_{k+1}=R$. So we have to find $A\in R$ such that \begin{cases}A\equiv a& \mkern-12mu\bmod J\\A\equiv a_{k+1},&\mkern-12mu\bmod I_{k+1}.\end{cases} which is possible by the case $k=2$, and the solution is unique modulo $J\cap I_{k+1} =\displaystyle\bigcap_{j=1}^{k+1} I_j$.