Why does a C.D.F need to be right-continuous? As you may know, if $(\Omega,\mathcal{F},\mathbb{P})$ is a probability space and $X\colon\Omega\to\mathbb{R}^k$ is a random variable, then the cumulative distribution function of $X$ is defined as 
$$F_X(a):=\mathbb{P}(\{\omega\in\Omega\,\colon X(\omega)\leq a\})=\mathbb{P}\Big(X^{-1}\Big(\prod_{i\in [k]}(-\infty,a_i]\Big)\Big),\,\text{ for each } a\in\mathbb{R}^k.$$
This function is always right-continuous. That is, for each $x\in\mathbb{R}^k$ we have $\lim_{a\downarrow x}F_X(a)=F_X(x)$. 
My question is: Why is this property important? Is there any capital result in probability theory that depends on it?

Well, in a finite measure (by which I mean a finite $\sigma$-additive measure) space, if $\{A_i\}_{i\in\Bbb N}$ is a sequence of measurable sets such that $A_i\supseteq A_{i+1}$ for all $i$, then $\mu\left(\bigcap_{n\in\Bbb N}A_i\right)=\inf_{n\in\Bbb N} \mu(A_i)=\lim_{n\to\infty} \mu(A_i)$. In your special case where all the $A_i$-s are hyperrectangle in the form $R\left(a^{(i)}\right)=\left(-\infty,a^{(i)}_1\right]\times\cdots\times\left(-\infty, a^{(i)}_k\right]$ and $\mu=\Bbb P_X$, this translates to $$\mathbb P_X\left(R(a)\right)=\mathbb P_X\left(\bigcap_{i\in\Bbb N} R\left(a^{(i)}\right)\right)=\lim_{n\to\infty} \mathbb P_X\left(R\left(a^{(i)}\right)\right)$$ for all $a^{(i)}\searrow a$. Which is in fact continuity on the right of the CDF.
