If strong duality holds, then it also holds when only a subset of the constraints is dualized.
We define the following three problems: the original, the partially dualized, and the dual.
Problem (P1):
\begin{align}\min_x&\quad f(x)\\\text{s.t.}&\quad g_i(x)\leq 0, i \in C\end{align}
Problem (P2):
\begin{align}\max_{\lambda\ge0} \min_x&\quad f(x) + \sum_{i \in A}\lambda_ig_i(x)\\\text{s.t.}&\quad g_i(x)\leq 0, i \in C\setminus A\end{align}
Problem (P3):
\begin{align}\max_{\mu\ge0} \max_{\lambda\ge0} \min_x&\quad f(x) + \sum_{i \in A}\lambda_ig_i(x) + \sum_{i \in C\setminus A}\mu_ig_i(x)\end{align}
It is given that strong duality holds, which means that (P1) and (P3) have the same objective value. For convenience, denote this by f(P1) = f(P3).
Using weak duality, we will show that f(P1) $\ge$ f(P2) $\ge$ f(P3). Because we know that f(P1) = f(P3), it must be that f(P1) = f(P2) = f(P3).
From (P1) to (P2): let $\bar{x}$ be an optimal solution to (P1). Because $\bar{x}$ is feasible for (P1), we have $g_i(\bar{x})\le0$ for all $i\in C$. Next, plug $\bar{x}$ into (P2), which is feasible. Due to the non-negativity of the multipliers, it follows for any $\lambda \ge 0$ that $f(\bar{x}) \ge f(\bar{x}) + \sum_{i \in A}\lambda_ig_i(\bar{x})$. Hence, f(P1) $\ge$ f(P2).
From (P2) to (P3): let $\bar{\lambda} \ge 0$ be optimal multipliers for (P2) and let $\bar{x}$ be corresponding optimal primal variables. Using a similar argument, $\bar{\lambda}$ and $\bar{x}$ can be plugged into (P3). Because $\mu \ge 0$ and $g_i(\bar{x})\le0$ for all $i\in C\setminus A$, we have for all $\mu \ge 0$ that
$$\quad f(\bar{x}) + \sum_{i \in A}\bar{\lambda}_ig_i(\bar{x}) \ge f(\bar{x}) + \sum_{i \in A}\bar{\lambda}_ig_i(\bar{x}) + \sum_{i \in C\setminus A}\mu_ig_i(\bar{x}).$$
It follows that f(P2) $\ge$ f(P3), which completes the proof.
