Random variable independent of itself

$\def\R{\mathbb R} \def\Z{\mathbb Z} \def\P{\mathbb P}$ As was already noticed in the comments the formulation is a bit misleading. Under the assumptions both sides of the equivalence must be true.

$X$ is independent of itself - in particular $\mathbb P(X\in B) = \mathbb P(X\in B, X \in B) = \mathbb P(X\in B) \cdot \mathbb P(X\in B) = \mathbb P(X\in B)^2$ (where $B$ is a borel subset of $\mathbb R$), which means that $\mathbb P(X\in B)$ is just equal to $0$ or $1$ for any $B$.

EDIT (more hints requested):

Now divide $\R$ into a contable number of disjoint intervals e.g. $I_n=[n,n+1)$ for $n\in \Z$. There exists a unique $n$ such that $\P(X\in I_n)=1$ (why?). Then you can divide $I_n$ into two halves and see which half has probability $1$. You can do it inductively and in the limit get a set of probability $1$ and diameter $0$. It must be a point and that's $a$ we were looking for.

Using the definition of independence one can see that for each $t$, we have $\mathbb P\{X\leqslant t\}\in\{0,1\}$. Define $$t_0:=\inf\{t\in\mathbb R\mid\mathbb P\{X\leqslant t\}=1\}.$$ The infimum exists since $\lim_{t\to +\infty}\mathbb P\{X\leqslant t\}=1$ (hence $\mathbb P\{X\leqslant t\}=1$ for $t$ large enough).

If $t_n\downarrow t_0$, then we can see that $\mathbb P\{X\leqslant t_0\}=1$ and since $\mathbb P\{X\leqslant s\}=0$ for $s\lt t_0$, we have $\mathbb P\{X\lt t_0\}=0$.

We conclude that $\mathbb P\left\{X=t_0\right\}=1$.