How do I show that the sum of two random variables is random variable?

There are a number of ways to do it. A standard trick for proving things like this is by noticing that $\{X+Y\leq x\}^c=\{X + Y > x\} = \displaystyle \bigcup_{r \in \mathbb{Q}} \{X > r\} \cap \{Y > x - r\} $, and that showing that this is in $F$ is enough to show that $X + Y$ is measurable. Then use the properties of $X$, $Y$, and $\sigma-$algebras to deduce that this set is measurable.

In fact, a random variable is a measurable function from $\Omega$ to $\mathbb{R}$.

\begin{align} & \{X+Y>x\}=\{X>x-Y\}=\bigcup_{q\in \mathbb{Q}}\{X>q>x-Y\} \\[10pt] = {} & \bigcup_{q\in \mathbb{Q}} (\{X>q\ \}\cap\{Y>x-q\}) = \bigcup_{q\in \mathbb{Q}}(\{X\le q\ \}^c \cap \{Y\le x-q\}^c) \end{align}

Since $ \{X\le q\ \}^c\in \mathcal{F}\ ,\ \{Y\le x-q\}^c\in \mathcal{F} $, $\mathbb{Q}$ is countable and $\mathcal{F}$ is a $\sigma$-field, we obtain $\{X+Y>x\}=\{X+Y\le x\}^c\in \mathcal{F}$. Then $$ \{\omega:X(\omega)+Y(\omega)\le x\}\in \mathcal{F}$$ $$ \{\omega:\min\{X(\omega),Y(\omega)\}\le x\}=\{\omega:X(\omega)\le x\}\bigcup\{\omega:Y(\omega)\le x\}\in \mathcal{F}$$ By definition, $X+Y$, and $\min\{X,Y\}$ are both random variables.