Suppose $x_1,x_2$ are solutions to:
\begin{align} x & \equiv c_1 \pmod{m_1} \\ & \,\,\,\vdots \\ x & \equiv c_k \pmod{m_k}\end{align}
Let's denote $M = \prod^k_{i=1} m_i$ and $y = x_1 - x_2$.
Hence, $$y\equiv 0 \pmod{m_i}\quad \forall i\in [k]$$
In other words, $$m_i \mid y \quad \forall i \in [k]$$
Since the $m_i$'s are co-prime then $$M\mid y \implies y\equiv 0 \pmod M$$
$$\implies x_1 \equiv x_2 \pmod{M}$$
I understand the proof, but, what if $M>y$? Then obviously $M\nmid y$
Or is it impossible? Why?