Optimization with parametric constraints: solution maps For constrained optimization problems
$$ \begin{array}{ll} \min\limits_{x \in \mathbb R^n} & f(p, x) \\
\text{s.t.} & x \in C \end{array} $$
where $p \in \mathbb R$ is a parameter, we can deduce, under suitable convexity conditions, existence and Lipschitzness (in $p$) of solution maps $x^*(p) \in \arg\min\limits_{x \in C} f(p, x)$, say, by applying Theorem 2F.10 from this Rockafellar's book. Here, I attached an excerpt from there:

Question: can we extend this result to the case when $C$ depends on $p$? Say, it is described by $g(p, x) \le 0$ with whatever properties required, like continuity in $p, x$, convexity in $x$ for each $p$.
The said book does not seem to contain a result in such a generic form.
One particular problem I see there is that the subgradients be pivoted to the static constraint. The argument collapses when the constraint changes

I'm almost sure that if the cost function is convex and the constraint doesn't change arbitrarily fast, the optimum must be continuous with a bounded rate of change
 A: This is not a direct answer, but also too long for a comment.
You may be interested in Berge's Maximum Theorem, which I quote below from Aliprantis and Border (2006) (Theorem 17.31). A correspondence $F\colon X \twoheadrightarrow Y$ is a set-valued function that maps each $x\in X$ into a subset $Y$. I will define continuity for correspondences after the statement of the theorem.

Theorem (Berge). Let $\varphi\colon X\twoheadrightarrow Y$ be a continuous correspondence between topological spaces with nonempty
  compact values, and suppose $f\colon X\times Y\to\mathbb R$ is
  continuous. Define the "value function" $m\colon X \to \mathbb R$ by
  $$ m(x) = \max_{y\in\varphi(x)} f(x,y), $$ and the correpsondence of
  maximisers by $$ \mu(x) = \{ y \in \varphi(x) : f(x,y) = m(x) \}. $$
  Then, 
  
  
*
  
*the value function is continuous;
  
*the argmax correspondence $\mu$ is upper hemicontinuous and has nonempty compact values.
  

The statement of this theorem in Aliprantis and Border is actually slightly more general than this (they require only continuity of $f$ on the graph of $\varphi$), but I have taken liberties for simplicity.
A neighbourhood of a set $A$ is any set $B$ containing an open set $V$ such that $A \subset V \subset B$.
We say that a correspondence $\varphi\colon X \twoheadrightarrow Y$ is upper hemicontinuous (uhc) if for every $x \in X$, and every neighbourhood $U$ of $\varphi (x)$, there is a neighbourhood $V$ of $x$ such that $z \in V$ implies that $\varphi (z) \subset U$.
A correspondence $\varphi\colon X \twoheadrightarrow Y$ is lower hemicontinuous (lhc) if for every $x \in X$ and every open set $U$ such that $\varphi (x) \cap U \neq \emptyset$, there is a neighbourhood $V$ of $x$ such that $\varphi (z) \cap U \neq \emptyset$ whenever $z \in V$.
A correspondence is continuous if it is both uhc and lhc.
One might wonder whether it is possible to find a continuous function $g$ such that $g(x) \in \mu(x)$ for each $x$, so that $g$ is a continuous selection from $\mu$. However, as we can see above, we can only guarantee that $\mu$ is uhc, and continuous selection theorems I am familiar with usually require that the correspondence we are selecting from be lhc.
However, since we know that $\mu$ is uhc, we also know that $\mu$ is continuous when viewed as a function from $X$ into $Y$ whenever we can guarantee that $\mu(x)$ is a singleton for each $x$. A sufficient condition for this is for $f$ to be strictly concave in $y$ for each $x$, with $\varphi$ being convex-valued (that is, $\varphi (x)$ is a convex set for each $x$).
To see this, suppose that $y,y^\prime \in \mu(x)$, with $y\neq y^\prime$. Then, from strict concavity of $f(\cdot ,x)$ and the fact that $\varphi(x)$ is convex, we have that, for $\alpha \in (0,1)$, $$ f(x, \alpha y + (1-\alpha) y^\prime) > \alpha f(x,y) + (1-\alpha)  f(x,y^\prime) = m(x)$$ and $\alpha y + (1-\alpha) y^\prime \in \varphi (x)$, a contradiction. This means that $\mu$ is uhc and singleton-valued, and is therefore continuous when viewed as a function.
