Let $X$ be a random variable that is independent of itself. I want to show that $\exists a \in \Bbb R: \Bbb P[X=a] = 1$.
Let $t \in \Bbb R$ be arbitrary. Then, $\Bbb P[X \le t] = \Bbb P[X \le t \cap X \le t] = (\Bbb P[X \le t])^2$, which implies that $\Bbb P[X \le t] \in \{0,1\}$. Especially there exists one $s \in \Bbb R: \Bbb P[X \le s] = 1 $.
Next, I decompose $\{X \le s\}$ into $\{\lfloor{s}\rfloor < X \le s\} \cup \bigcup_{k=1}^{\infty} \{\lfloor{s}\rfloor -(k+1) < X \le \lfloor{s}\rfloor -k\}$ and since they are disjoint only one of these sets has probability $1$. Continuing this idea, we get a monotone decreasing sequence of sets $(A_n)_{n \in \Bbb N}$ with $\Bbb P[A_n] = 1$ for all $n \in \Bbb N$ and $\bigcap_{n \in \Bbb N} A_n = \{X=a\}$ for some $a \in \Bbb R$. From the continuity of the measure we obtain $\Bbb P[X=a]$ =1.
Assume, there exists another $a' \in \Bbb R: \Bbb P[X=a'] = 1$. Then $\Bbb P[X = a \cup X = a'] = \Bbb P[X=a] + \Bbb P[X=a'] = 2$ leading to a contradiction.
Let $F$ be the c.d.f. of $X$ (i.e., $F(x)=P\left([X\leq x]\right)$. As pointed out by you, for each $x$, we have $F(x)=0$ or $F(x)=1$. Let $A=\{x\in\mathbb{R}\mid F(x)=1\}$. Note that $A$ is non-empty. (For, if $A$ is empty, then $F(x)=0$ for all $x$, which contradicts to the property $\lim_{x\rightarrow\infty}F(x)=1$.) Since $\lim_{x\rightarrow-\infty}F(x)=0$, by the same reasoning, $A$ is bounded from below. Let $a=\inf A\in\mathbb{R}$, then there exists a sequence $(x_{n})$ in $A$ such that $x_{1}\geq x_{2}\geq\ldots\geq a$ and $x_{n}\rightarrow a$. By the right-continuity of $F$, we have $F(a)=\lim_{n\rightarrow\infty}F(x_{n})=1$. For any $x<a,$ we have $x\notin A$, so $F(x)=0$. Hence, $F(a-)=\lim_{x\rightarrow a-}F(x)=0$. Finally, $P\left([X=a]\right)=F(a)-F(a-)=1$.