If $X$ and $Y$ are random variables, is $Z=(X, Y)$ a random variable? What definition do we use for random variable? How do we show that $Z=(X,Y)$ is a random variable?
 A: We would call $Z$ a random vector (but yes, it is random).
Random variables are defined within a probability space that includes a sample space $\Omega$.  Within such a space, real-valued random variables $X:\Omega \rightarrow \mathbb{R}$ and $Y:\Omega \rightarrow \mathbb{R}$ are functions$^\dagger$ from the sample space to Euclidean space.  You have $Z \equiv (X,Y)$ so $Z(\omega) = (X(\omega), Y(\omega)) \in \mathbb{R}^2$ for all $\omega \in \Omega$, which means that the latter is a mapping $Z:\Omega \rightarrow \mathbb{R}^2$.$^\dagger$  This means that it is a two-element real random vector.

$^\dagger$ Formally, random variables must be measurable functions, which is a formal requirement from measure theory.  If you are new to dealing with random variables, just ignore this part; I mention it for completeness.  It can be shown that measureability of $X$ and $Y$ implies measureability of $Z$, so this requirement is met in this case.  (Hat-tip to whuber in the comments for noting the importance of the latter property.)
A: To be formal, you need to set up a probability space $(\Omega, \mathscr{F}, P)$ (to discuss the measurability only, the probability measure $P$ is even not required, as the argument below shows). What you want to confirm is that given $X$ and $Y$ are two $\mathscr{F}$-measurable mappings from $\Omega$ to $\mathbb{R}^1$, then $Z$ is an $\mathscr{F}$-measurable mapping from $\Omega$ to $\mathbb{R}^2$. 
This is ready to check by noting that for any $(x, y) \in \mathbb{R}^2$, we have 
\begin{align}
& \{\omega: Z(\omega) \leq (x, y)\} \\
= & \{\omega: X(\omega) \leq x, Y(\omega) \leq y\} \\
= & \{\omega: X(\omega) \leq x\} \cap \{\omega: Y(\omega) \leq y\} \in \mathscr{F}
\end{align}
as both sets after the last equality sign are in $\mathscr{F}$ by measurability of $X$ and $Y$ and that the $\sigma$-field $\mathscr{F}$ is closed under intersection. 
Note also the converse of the statement is also true: if $Z$ is a random vector, then each of its component is a random variable. 
