[proofplan]
The interval identity follows by writing $\{a<X\le b\}$ as the set difference between the events $\{X\le b\}$ and $\{X\le a\}$. The point-mass identity follows by writing $\{X=a\}$ as the difference between $\{X\le a\}$ and $\{X<a\}$. The only limiting point is to identify $\mathbb P(X<a)$ with the left-hand limit of the distribution function, which is obtained from continuity from below for an increasing sequence of events.
[/proofplan]
[step:Declare the probability space and the distribution function]
Let $(\Omega,\mathcal F,\mathbb P)$ be the probability space. Let $\mathcal B(\mathbb R)$ denote the Borel $\sigma$-algebra on $\mathbb R$, and let
\begin{align*}
X:(\Omega,\mathcal F)&\to(\mathbb R,\mathcal B(\mathbb R))
\end{align*}
be the real-valued random variable. Define the distribution function
\begin{align*}
F_X:\mathbb R&\to[0,1]\\
x&\mapsto \mathbb P(\{\omega\in\Omega:X(\omega)\le x\}).
\end{align*}
Since $(-\infty,x]$, $(a,b]$, $(-\infty,a)$, and $\{a\}$ are Borel subsets of $\mathbb R$, the events obtained by taking their preimages under the measurable map $X$ belong to $\mathcal F$.
[/step]
[step:Subtract nested distribution events to compute $\mathbb P(a<X\le b)$]
Fix real numbers $a,b\in\mathbb R$ with $a<b$. Define the events
\begin{align*}
A&:=\{\omega\in\Omega:X(\omega)\le a\},\\
B&:=\{\omega\in\Omega:X(\omega)\le b\}.
\end{align*}
Because $a<b$, we have $A\subset B$, and
\begin{align*}
B\setminus A=\{\omega\in\Omega:a<X(\omega)\le b\}.
\end{align*}
Finite additivity of the probability measure $\mathbb P$ on the disjoint union $B=A\cup(B\setminus A)$ gives
\begin{align*}
F_X(b)=\mathbb P(B)=\mathbb P(A)+\mathbb P(B\setminus A)=F_X(a)+\mathbb P(\{\omega\in\Omega:a<X(\omega)\le b\}).
\end{align*}
Subtracting $F_X(a)$ gives
\begin{align*}
\mathbb P(\{\omega\in\Omega:a<X(\omega)\le b\})=F_X(b)-F_X(a).
\end{align*}
[/step]
[step:Identify $\mathbb P(X<a)$ with the left limit of $F_X$]
Fix a real number $a\in\mathbb R$. Let $\mathbb N:=\{1,2,3,\ldots\}$. For each $n\in\mathbb N$, define the event
\begin{align*}
C_n:=\{\omega\in\Omega:X(\omega)\le a-\frac{1}{n}\}.
\end{align*}
The sequence $(C_n)_{n\in\mathbb N}$ is increasing, because $a-\frac{1}{n}\le a-\frac{1}{n+1}$. Its union is
\begin{align*}
\bigcup_{n\in\mathbb N} C_n=\{\omega\in\Omega:X(\omega)<a\}.
\end{align*}
Indeed, if $X(\omega)\le a-\frac{1}{n}$ for some $n\in\mathbb N$, then $X(\omega)<a$; conversely, if $X(\omega)<a$, then $a-X(\omega)>0$, and by the Archimedean property there exists $n\in\mathbb N$ such that $\frac{1}{n}<a-X(\omega)$, hence $X(\omega)<a-\frac{1}{n}\le a-\frac{1}{n+1}$.
By continuity from below for probability measures applied to the increasing sequence $(C_n)_{n\in\mathbb N}$,
\begin{align*}
\mathbb P(\{\omega\in\Omega:X(\omega)<a\})
=\lim_{n\to\infty}\mathbb P(C_n)
=\lim_{n\to\infty}F_X(a-\frac{1}{n}).
\end{align*}
Since $F_X$ is nondecreasing, the left-hand limit exists and equals
\begin{align*}
\lim_{x\uparrow a}F_X(x)=\sup_{x<a}F_X(x)=\lim_{n\to\infty}F_X(a-\frac{1}{n}).
\end{align*}
Therefore
\begin{align*}
\mathbb P(\{\omega\in\Omega:X(\omega)<a\})=\lim_{x\uparrow a}F_X(x).
\end{align*}
[guided]
We want to convert the strict event $\{X<a\}$ into distribution-function values, but $F_X$ directly measures only events of the form $\{X\le x\}$. The standard way to approximate $\{X<a\}$ from below is to use levels that increase toward $a$.
Fix $a\in\mathbb R$, and define $\mathbb N:=\{1,2,3,\ldots\}$. For each $n\in\mathbb N$, define
\begin{align*}
C_n:=\{\omega\in\Omega:X(\omega)\le a-\frac{1}{n}\}.
\end{align*}
Each $C_n$ is an event in $\mathcal F$, because $(-\infty,a-\frac{1}{n}]$ is a Borel subset of $\mathbb R$ and $X:(\Omega,\mathcal F)\to(\mathbb R,\mathcal B(\mathbb R))$ is measurable. The sequence is increasing: since
\begin{align*}
a-\frac{1}{n}\le a-\frac{1}{n+1},
\end{align*}
the implication $X(\omega)\le a-\frac{1}{n}\implies X(\omega)\le a-\frac{1}{n+1}$ gives $C_n\subset C_{n+1}$.
Now identify the union. If $\omega\in\bigcup_{n\in\mathbb N}C_n$, then $\omega\in C_n$ for some $n\in\mathbb N$, so
\begin{align*}
X(\omega)\le a-\frac{1}{n}<a.
\end{align*}
Thus $\omega\in\{\omega\in\Omega:X(\omega)<a\}$. Conversely, suppose $\omega\in\Omega$ satisfies $X(\omega)<a$. Then $a-X(\omega)>0$. By the Archimedean property of the real numbers, there exists $n\in\mathbb N$ such that
\begin{align*}
\frac{1}{n}<a-X(\omega).
\end{align*}
Equivalently,
\begin{align*}
X(\omega)<a-\frac{1}{n}.
\end{align*}
Hence $X(\omega)\le a-\frac{1}{n}$, so $\omega\in C_n$. Therefore
\begin{align*}
\bigcup_{n\in\mathbb N}C_n=\{\omega\in\Omega:X(\omega)<a\}.
\end{align*}
Because $(C_n)_{n\in\mathbb N}$ is increasing and $\mathbb P$ is a probability measure, continuity from below for measures applies. Its hypotheses are satisfied: each $C_n$ belongs to $\mathcal F$, the inclusions $C_n\subset C_{n+1}$ hold, and $\mathbb P$ is countably additive. Therefore
\begin{align*}
\mathbb P(\{\omega\in\Omega:X(\omega)<a\})
=\mathbb P\left(\bigcup_{n\in\mathbb N}C_n\right)
=\lim_{n\to\infty}\mathbb P(C_n).
\end{align*}
By the definition of $F_X$,
\begin{align*}
\mathbb P(C_n)=F_X(a-\frac{1}{n}),
\end{align*}
so
\begin{align*}
\mathbb P(\{\omega\in\Omega:X(\omega)<a\})
=\lim_{n\to\infty}F_X(a-\frac{1}{n}).
\end{align*}
It remains to connect this sequential limit with the left-hand limit $\lim_{x\uparrow a}F_X(x)$. The distribution function $F_X$ is nondecreasing because, whenever $x\le y$, the event $\{X\le x\}$ is contained in the event $\{X\le y\}$, and monotonicity of $\mathbb P$ gives
\begin{align*}
F_X(x)\le F_X(y).
\end{align*}
Hence the left-hand limit exists as the supremum of the values below $a$:
\begin{align*}
\lim_{x\uparrow a}F_X(x)=\sup_{x<a}F_X(x).
\end{align*}
Since $a-\frac{1}{n}<a$ for every $n\in\mathbb N$, we have
\begin{align*}
\lim_{n\to\infty}F_X(a-\frac{1}{n})\le \sup_{x<a}F_X(x).
\end{align*}
For the reverse inequality, take any real number $x<a$. By the Archimedean property, there exists $n\in\mathbb N$ such that $x\le a-\frac{1}{n}$. Since $F_X$ is nondecreasing,
\begin{align*}
F_X(x)\le F_X(a-\frac{1}{n})\le \lim_{m\to\infty}F_X(a-\frac{1}{m}).
\end{align*}
Taking the supremum over all $x<a$ gives
\begin{align*}
\sup_{x<a}F_X(x)\le \lim_{m\to\infty}F_X(a-\frac{1}{m}).
\end{align*}
Thus
\begin{align*}
\lim_{x\uparrow a}F_X(x)=\lim_{n\to\infty}F_X(a-\frac{1}{n}).
\end{align*}
Combining this with the continuity-from-below computation yields
\begin{align*}
\mathbb P(\{\omega\in\Omega:X(\omega)<a\})=\lim_{x\uparrow a}F_X(x).
\end{align*}
[/guided]
[/step]
[step:Subtract the strict lower event to compute the atom at $a$]
Fix a real number $a\in\mathbb R$. Define the events
\begin{align*}
D&:=\{\omega\in\Omega:X(\omega)<a\},\\
E&:=\{\omega\in\Omega:X(\omega)\le a\}.
\end{align*}
Then $D\subset E$ and
\begin{align*}
E\setminus D=\{\omega\in\Omega:X(\omega)=a\}.
\end{align*}
By finite additivity on the disjoint union $E=D\cup(E\setminus D)$,
\begin{align*}
F_X(a)=\mathbb P(E)=\mathbb P(D)+\mathbb P(E\setminus D)
=\lim_{x\uparrow a}F_X(x)+\mathbb P(\{\omega\in\Omega:X(\omega)=a\}).
\end{align*}
Subtracting the left-hand limit gives
\begin{align*}
\mathbb P(\{\omega\in\Omega:X(\omega)=a\})=F_X(a)-\lim_{x\uparrow a}F_X(x).
\end{align*}
Together with the interval identity proved above, this is exactly the desired result.
[/step]
