$\Theta \rho ϵ\eta \Pi \alpha \tau \pi$

Let $F=F_X$. If $x\le y$, then every $\omega \in \{X\le x\}$ satisfies $X\left(\omega \right)\le x\le y$, so $\omega \in \{X\le y\}$. Thus $\{X\le x\}\subseteq \{X\le y\}$, and monotonicity of probability gives $$F\left(x\right)=P(X\le x)\le P(X\le y)=F\left(y\right).$$ Therefore $F$ is non-decreasing.

For the limit at infinity, define $A_n:=\{X\le n\}$. Then $A_n\nearrow \mathrm{\Omega }$, because every real value $X\left(\omega \right)$ is at most some integer $n$. By continuity of probability,

$$\underset{n\to \infty }{\lim }F\left(n\right)=\underset{n\to \infty }{\lim }P\left(A_n\right)=P\left(\mathrm{\Omega }\right)=1.$$ Since $F$ is non-decreasing and $F\left(x\right)\le 1$, integer values determine the monotone function limit at infinity gives ${\lim }_{x\to \infty }F\left(x\right)=1$.

For the limit at negative infinity, define $B_n:=\{X\le -n\}$. Then $B_n\searrow \varnothing$: the events are decreasing, and no real number is less than or equal to $-n$ for every $n$. By continuity of probability and probability of the impossible event,

$$\underset{n\to \infty }{\lim }F(-n)=\underset{n\to \infty }{\lim }P\left(B_n\right)=P\left(\varnothing \right)=0.$$ Since $F$ is non-decreasing and $F\left(x\right)\ge 0$, this implies ${\lim }_{x\to -\infty }F\left(x\right)=0$.

Finally, fix $x\in ℝ$ and define $C_n:=\{X\le x+\frac{1}{n}\}$. Then $C_n\searrow \{X\le x\}$: the events are decreasing, and $\omega$ lies in every $C_n$ exactly when $X\left(\omega \right)\le x$. By continuity of probability,

$$\underset{n\to \infty }{\lim }F(x+\frac{1}{n})=\underset{n\to \infty }{\lim }P\left(C_n\right)=P(X\le x)=F\left(x\right).$$ Since $F$ is non-decreasing, the values $F(x+h)$ with $h\to 0^{+}$ are squeezed between $F\left(x\right)$ and values of the form $F(x+\frac{1}{n})$ with $n\to \infty$. Hence ${\lim }_{h\to 0^{+}}F(x+h)=F\left(x\right)$, so $F$ is right-continuous.

Let $$E_a:=\{X\le a\}\text{and}{E}_b:=\{X\le b\}.$$ Since $a<b$, if $\omega \in E_a$, then $X\left(\omega \right)\le a<b$, so $\omega \in E_b$. Thus $E_a\subseteq E_b$.

By set difference decomposition, $E_b$ is the disjoint union of $E_b\setminus E_a$ and $E_b\cap E_a$. Also,

$$E_b\setminus E_a=\{a<X\le b\},$$ because $\omega \in E_b\setminus E_a$ exactly when $X\left(\omega \right)\le b$ and it is not true that $X\left(\omega \right)\le a$, which is the same as $a<X\left(\omega \right)\le b$. Since $E_a\subseteq E_b$, intersection with a superset is itself gives $E_a\cap E_b=E_a$.

Therefore, by probability of set difference,

$$P(a<X\le b)=P(E_b\setminus E_a)=P\left(E_b\right)-P(E_b\cap E_a)=P\left(E_b\right)-P\left(E_a\right)=F_X\left(b\right)-F_X\left(a\right).$$