Why does the cumulative distribution function for discrete random variable right continuous? If the random variable is discrete, then the cumulative value should also be discrete because the variable can only take on discrete values, right?
 A: No
The cumulative distribution function for a random variable $X$ supported on some subset of the real numbers can be defined as $$F(x)= \mathbb P(X \le x)$$ for all real $x$ whether $X$ is a discrete random variable or a continuous random variable or some mixture.  $F(x)$ is an increasing (perhaps weakly increasing) function of $x$ with $\lim\limits_{x \to -\infty}F(x) = 0$ and $\lim\limits_{x \to +\infty}F(x) = 1$.  
What is special about random variables which are wholly or partly discrete, i.e. with some $y$ where $\mathbb P(X = y)>0$ is that $F(x)$ is left-discontinuous at $y$ since $$F(y)-\lim\limits_{x \to y^{-}} F(x)= \mathbb P(X \le y) - \mathbb P(X \lt y) = \mathbb P(X = y)>0$$ 
But even here $F(x)$ is right-continuous at $y$ since $$\lim\limits_{x \to y^{+}} F(x) - F(y)= \lim\limits_{x \to y^{+}}\mathbb P(X \le x) - \mathbb P(X \le y) =\mathbb P(X \le y) - \mathbb P(X \le y) = 0$$
As an example, suppose $X$ takes the values $\pm3$ with equal probability so $\mathbb P(X =-3) = \mathbb P(X =3) =\frac12$. Then 
$$F(x)=\mathbb P(X \le x) =\begin{cases} 
0 & \text{ when }x \lt 3\\
\frac12 & \text{ when }-3 \le x \lt 3\\
1 & \text{ when }3 \le x \\
\end{cases}$$
which is clearly right-continuous at $\pm3$ and also everywhere else.
