I was self-reading Section 2.7 of Karatzas and Shreve's Brownian Motion and Stochastic Calculus and Proposition 7.7 claims that the Strong Markov processes augmented filtration is right continuous. In the proof(screenshots below), according to me only the Markov property is being used since we consider only a deterministic optional time $s$ in the proof.
I understand most of the details in the proof but it seems I don't really see where exactly is the strong Markov property is being used?

I am at a total loss and I would really appreciate if someone could help me here.
You are right that the strong Markov property is not used in its full generality; somehow, it's an artifact of the definition of the (strong) Markov property by Karatzas & Shreve. They say that $(X_t,\mathcal{F}_t)_t$ is Markov if
$$\mathbb{E}(f(X_{s+t}) \mid \mathcal{F}_s) = \mathbb{E}(f(X_{s+t}) \mid X_s) \tag{1}$$
and strong Markov if
$$\mathbb{E}(f(X_{S+t}) \mid \mathcal{F}_{S+}) = \mathbb{E}(f(X_{S+t}) \mid X_S) \tag{2}$$
for any optional time $S$. Clearly, $(2)$ implies
$$\mathbb{E}(f(X_{s+t}) \mid \mathcal{F}_{s+}) = \mathbb{E}(f(X_{s+t}) \mid X_s) \tag{3}$$
Note that $(3)$ differs from $(1)$ since we are conditioning on $\mathcal{F}_s$ and $\mathcal{F}_{s+}$, respectively. It follows easily from the tower property that $(3)$ implies $(1)$ but the converse is not immediate. Karatzes-Shreve need $(3)$ for the proof of the right-continuity of the filtration, and therefore they are refering to the strong Markov property.
The proof by Karatzas-Shreve actually shows the following statement (which can be, for instance, also found in the book by Blumenthal & Getoor)