The dozens of proofs I've seen for showing that a tail sigma-algebra measurable random variable $X$ is almost surely a constant use the infimum $T := \inf \{t \in \mathbb{R}\mid \mathbb{P}[\{X \leq t\} = 1\}$, and state that: "[...] by the right continuity of the cumulative distribution function it follows that $\mathbb{P}[\{X \leq T\}] = 1$ and $\mathbb{P}[\{X < T\}] = 0$".
I have to say that it is not quite clear to me the events $\{X \leq T\}$ and $\{X < T\}$ have the stated probabilities.
Is the following equivalent to the hidden reasoning behind the claimed value $\mathbb{P}[\{X \leq T\}] = 1$?
Let $t_n = T + \frac{1}{n}$. By right continuity of the CDF we know that if $t_n\downarrow t$, then $F(t_n)\downarrow F(t)$.Then $t_n\downarrow T$, and thus $1 = F(T) \leq F\left(T + \frac{1}{n}\right)\downarrow F(T) = 1$, where I suppose $F(T) = 1$ by definition, even though we are giving the CDF the infimum of a set, so that isn't it possible for $T = \pm \infty$? As for the $\mathbb{P}[\{X < T\}] = 0$, I get that the idea is that as $T$ is the smallest real number (0r $\pm\infty$?) for which $\mathbb{P}[\{X \leq T\}] = 1$ and $F$ takes only the values of $0$ or $1$, it then follows that for any $t < T: F(t) \neq 1 \implies F(t) = 0$. But I don't know how to show this with some real sequence.
You are right that $T$ might be $\infty$ or $-\infty$. These exceptional cases are easily handled. So assume that $T \in \mathbb{R}$. $F(T) = 1$ is not true by definition. What you know by definition is that if $t < T$, then $F(t) = 0$. To get the full graph of $F$, it remains to prove that $F(T) = 1$. By definition of $\inf$, for any $n > 0$, there exists $t < T + 1/n$ such that $F(t) = 1$. Thus $F(T + 1/n) = 1$ for all $n > 0$. Now letting $n \to \infty$ and using right continuity of $F$ gives $F(T) = 1$.
Since $P(X < T) = \lim_{t \nearrow T}F(t) = 0$, it follows that $P(X = t) = 1$.