It's an immediate consequence of uniqueness of solutions in CRT (Chinese Remainder Theorem). The proof is quite easy. $\:$ If $\rm\:gcd(m,n)\ =\ 1\:$ then $\rm\ x\ mod\ mn\ \to\ (x\ mod\ m,\ x\ mod\ n)\ $ is $1$ to $1\:.\ $ For if $\rm\ x,\:y\ $ map to the same value then $\rm\: x = y\ (mod\ m)\: $ so $\rm\ m\ |\ x-y\:,\:$ i.e. $\rm\:m\:$ divides $\rm\:x-y\:.\:$ Similarly $\rm\:n\ |\ x-y\:.\:$ Hence $\rm\: lcm(m,n)\ |\ x-y\:.\:$ But $\rm\:gcd(m,n) = 1\ \iff\ lcm(m,n) = mn\:.\:$ Thus $\rm\ mn = lcm(m,n)\ |\ x-y\:,\:$ hence $\rm\ x = y\ (mod\ mn)\:.\:$ Therefore, indeed, the map is $1$ to $1$.
Conversely, if $\rm\ lcm(m,n) < mn\ $ then $\rm\:x\:$ and $\rm\ x + lcm(m,n)\ $ are not congruent $\rm\:(mod\ mn)\:.\:$ However, both map to the same value $\rm\ (x\ mod\ m,\ x\ mod\ n)\:,\ $ since both $\rm\: m\:$ and $\rm\:n\ |\ lcm(m,n)\:.\ $ This explains the example you gave in your comment, where $\rm\:m,n = 6,10\:$ so $\rm\:lcm(m,n) = 30\:,\:$ hence $\rm\: x= 0\:$ and $\rm\:x+30 = 30\:$ both map to $\rm\:(0,0)\:$ but $\rm\: 30\not\equiv 0\ (mod\ 60)\:.$
The existence of solutions is also easy. Clearly $\rm\:|\mathbb Z\ mod\ mn| = mn = |\mathbb Z\ mod\ m \times \mathbb Z\ mod\ n|\: $ so the above map being $1$ to $1$ on these equal cardinality finite sets implies the map is also onto, i.e. every pair $\rm\:(i\ mod\ m,\:j\ mod\ n)\:$ is the image of some $\rm\: x\ mod\ mn\:.\:$ Summing up: if $\rm\:gcd(m,n) = 1$ then there exists a unique solution $\rm\ x\ mod\ mn\ $ to the congruences $\rm\ x\equiv i\ (mod\ m),\ x\equiv j\ (mod\ n)\:.\ $ For a simple constructive proof see here.