7. Proof: From $x \equiv a\left(\bmod \left\{m_{1}, m_{2}\right\}\right)$, we know
$$\left\{m_{1}, m_{2}\right\} \mid(x-a)$$

Since $m_{1}\left|\left\{m_{1}, m_{2}\right\}, m_{2}\right|\left\{m_{1}, m_{2}\right\}$, by Lemma 2 of Chapter 1, we have
$$\begin{array}{c} m_{1}\left|(x-a), \quad m_{2}\right|(x-a), \\ x \equiv a \quad\left(\bmod m_{1}\right), \quad x \equiv a \quad\left(\bmod m_{2}\right) \end{array}$$

This proves that $x \equiv a\left(\bmod \left\{m_{1}, m_{2}\right\}\right)$ is a solution to the system of congruences $x \equiv a\left(\bmod m_{1}\right)$, $x \equiv a\left(\bmod m_{2}\right)$.

Conversely, suppose $x_{0}$ is a solution to $x \equiv a\left(\bmod m_{1}\right), x \equiv a\left(\bmod m_{2}\right)$. By the definition of congruence, we have
$$m_{1}\left|\left(x_{0}-a\right), \quad m_{2}\right|\left(x_{0}-a\right)$$

Therefore, by Lemma 9 of Chapter 1, $\left\{m_{1}, m_{2}\right\} \mid\left(x_{0}-a\right)$, i.e.,
$$x_{0} \equiv a\left(\bmod \left\{m_{1}, m_{2}\right\}\right)$$

Thus, $x \equiv a\left(\bmod \left\{m_{1}, m_{2}\right\}\right)$ is the complete solution to the system of congruences $x \equiv a\left(\bmod m_{1}\right)$, $x \equiv a\left(\bmod m_{2}\right)$.