Let $X$ be a stochastic process in continuous time and $T$ be a stopping time.
All the standard texts I've looked at so far, as well as this excellent blog, introduce the assumption that $X$ is progressively measurable to prove that $X_T$ is $\mathcal F_T$-measurable. Why is merely adapted $X$ not enough? Indeed the proof in the linked article seems to use "progressive" solely to claim "adapted" for the conclusion.
The proof in the article uses progressive to claim that the stopped process $X^T$ is adapted, not that $X$ is adapted. The reason that something stronger than adaptedness is needed is because $X_T$ may fail to even be $\mathcal F_\infty$ measurable without further assumptions.
As an example, consider the probability space $\Omega := [0,1]$ equipped with the Lebesgue measure and the Borel (or Lebesgue) $\sigma$-algebra $\mathcal F$. We will use the constant filtration $\mathcal F_t = \mathcal F$ for all $t$. Define the stopping time $T$ by $T(\omega) = \omega$ for all $\omega \in \Omega$.
Let $A \subset [0,1]$ be any subset which is not $\mathcal F$ measurable, and define the stochastic process $X$ by $X_t(\omega) = 1_{\omega = t, t \in A}$. Observe that, for all $t$, $$\mathbb{P}({X_t \ne 0}) = \mathbb{P}(\omega = t, t \in A) \le \mathbb{P}(\omega = t) = 0, $$ so $X_t = 0$ almost surely. In particular $X_t$ is $\mathcal F_t$ measurable, i.e. $X$ is adapted. However, $$ X_{T(\omega)}(\omega) = 1_{\omega = T(\omega),T(\omega) \in A} = 1_{\omega = \omega, \omega \in A} = 1_{\omega \in A}, $$ which is not $\mathcal F$ measurable by the definition of $A$. Hence $X_T$ is not $\mathcal F$ measurable, and in particular is not $\mathcal F_T$ measurable.