The jumping times of a càdlàg process are stopping times.
The key point is the following lemma which has nothing to do with probability theory, but is just elementary "deterministic" calculus.
Lemma: Let $f: [0,\infty) \to \mathbb{R}$ be a càdlàg function and fix $M>0$ and $0 \leq T_0<T_1<\infty$. Then the following statements are equivalent:
There exists $t \in (T_0,T_1]$ such that $|\Delta f(t)|>M$.
There exists $K \in \mathbb{N}$ such that for any $k \in \mathbb{N}$ there are $u,v \in \mathbb{Q} \cap (T_0,T_1+k^{-1}]$ such that $|u-v| \leq k^{-1}$ and $|f(u)-f(v)|>M+\frac{1}{K}$.
Applying this lemma with $T_0 =0$ and $T_1 = t$ and $f(t) := X_t(\omega)$, we find
$$\{T^1\leq t\} = \bigcup_{K \geq 1} \bigcap_{k \in \mathbb{N}} \bigcup_{\substack{u,v \in \mathbb{Q} \cap (0,t+k^{-1}) \\ |u-v| \leq k^{-1}}} \{|X_u-X_v| > M+1/K\}.$$
Since the right-hand side is a countable union/intersection, this shows $\{T^1 \leq t\} \in \mathcal{F}_{t+}$ and so $T^1$ is an $(\mathcal{F}_{t+})$-stopping time. If the filtration is right-continuous, then $T^1$ is an $(\mathcal{F}_t)_t$-stopping time.
Similarly, it holds that
$$\{T^{n+1} \leq t\} = \bigcup_{q \in \mathbb{Q} \cap [0,t)} \{T^n \leq q\} \cap \bigg(\bigcup_{K \geq 1} \bigcap_{k \in \mathbb{N}} \bigcup_{\substack{u,v \in \mathbb{Q} \cap (q,t+k^{-1}) \\ |u-v| \leq k^{-1}}} \{|X_u-X_v|>M+1/K\} \bigg).$$
Again, this is a countable union/intersection and therefore $T^{n+1}$ is an $(\mathcal{F}_{t+})$-stopping time.