You have stated Troelstra's Uniformithy Principle correctly (contrary to François's claim). There are two reasons why your counterexample does not work. First, we cannot show that every $X \subseteq \mathbb{N}$ is empty or not. Second, we cannot show that every inhabited subset of $\mathbb{N}$ has a minimal element. And we do not need any story about constructivism here. As soon as we assume the Uniformity Principle, it follows that:

Not every $X \subseteq \mathbb{N}$ is empty or not, because that would allows us to form the non-uniform total relation $(X = \emptyset \implies n = 1) \land (X \neq \emptyset \implies n = 42)$.

Not every inhabited subset $X \subseteq \mathbb{N}$ has a minimal element, because that would allow us to form the non-uniform total relation $\min (X \cup \lbrace 42 \rbrace) = n$.

[I added this paragraph later.] Even without the uniformity principle we cannot expect the two statements to be constructively valid because they imply (a restricted form of) excludded middle:

If every set is empty or not, given any truth value $p \in \Omega$, consider the set $\lbrace \star \mid p \rbrace$: it is either empty or not, therefore either $\lnot p$ or $\lnot\lnot p$, which is a restricted form of excluded middle. [EDIT: thanks to Andreas Blass for pointing out an error here.]

If every inhabited set of natural numbers has a minimum, given any truth value $p \in \Omega$, the minimum of the set $\lbrace n \in \mathbb{N} \mid n > 1 \lor p\rbrace$ is either $0$ or not, therfore either $p$ or $\lnot p$, which is excluded middle.

There is no hope to make the Uniformity Principle classically valid, even if we try to restrict to a subfamily of sets:

**Theorem:** Suppose excluded middle holds and $S \subseteq \mathcal{P}(\mathbb{N})$ is an inhabited family which satisfies the *uniformity principle* $(\forall X \in S \exists n \in \mathbb{N} . R(X,n)) \implies \exists n \in \mathbb{N} \forall X \in S . R(X, n)$. Then $S$ contains exactly one set.

Proof. Easy exercise. But let's do it anyway. There is an inhabitant $X_0 \in S$. Define the map $f : S \to \mathbb{N}$ by (use of excluded middle coming up...)
$$f(X) = \begin{cases}
42 & \text{if $X = X_0$} \\\\
23 & \text{if $X \neq X_0$}
\end{cases}$$
Let $R(X,n)$ be the relation $f(X) = n$. Since $n = 42$ is the only $n$ related to $X_0$, by Uniformity Principle for $S$ we have $\forall X \in S . f(X) = 42$, therefore $\forall X \in S . X = X_0$. QED.

That is, Troelstra's Uniformity Principle is a very non-classical axiom. I disagree that it is a continuity principle. It does not allow one to prove any of the usual continuity statements "all maps are continuous".