Given the Pauli matrices $X$ and $Z$ defined as
$$X = \begin{pmatrix}0 & 1\\ 1& 0 \end{pmatrix},\ \ Z = \begin{pmatrix}1 & 0\\ 0& -1 \end{pmatrix}$$
I read a result (see page 2) where it is claimed that
any 1-qubit gate can be realized as a composition of $e^{i\theta_1Z}e^{i\theta_2 X}e^{i\theta_3 Z}$, according to the Euler's formula
I believe this is equivalent to saying any $2\times 2$ unitary can be written in this form. How does one show this?
There is no global phase in the equation you cited, so any special unitary $2 \times 2$ matrix can be written in this form. Global phase is often disregarded in quantum computations, so it's not unusual that the authors would leave it out.
The proof is straightforward, since $U \in SU(2)$, the rows and columns of $U$ are orthonormal, and the determinant of $U$ is $1$, from which it follows that there exist real numbers $\theta_1, \theta_2, \theta_3$ such that $$U=\begin{bmatrix} e^{i \theta_1} & 0 \\ 0 & e^{-i \theta_1} \end{bmatrix} \begin{bmatrix} \cos \theta_2 & i \, \sin \theta_2 \\ i \, \sin \theta_2 & \cos \theta_2 \end{bmatrix} \begin{bmatrix} e^{i \theta_3} & 0 \\ 0 & e^{-i \theta_3} \end{bmatrix}= \begin{bmatrix} e^{i(\theta_1 + \theta_3)} \cos \theta_2 & i\, e^{i(\theta_1-\theta_3)} \sin \theta_2 \\ i \, e^{i(-\theta_1+\theta_3)} \sin \theta_2 & e^{i(-\theta_1 - \theta_3)} \cos \theta_2 \end{bmatrix}.$$ This is only a minor change from the proof of Theorem 4.1 in Mike and Ike for Z-Y decomposition. If it is still fuzzy for you, I would suggest reading through this section (4.2) carefully.