Consider the following modification of the Dirichlet "popcorn" function: $$f(x) = \begin{cases} 1/q, & \text{$x \in \mathbb{Q}$, $x=p/q$ in lowest terms} \\ -1, & x \notin \mathbb{Q},\, x < 0 \\ -2, & x \notin \mathbb{Q}, \, x > 0.\end{cases}$$ Since every real number can be approximated by rationals with arbitrarily large denominator, the graph of $f$ contains the $x$-axis, which is the uniformly continuous function $0$.

Let $S \subset \mathbb{R}$ be dense. If $f|_S$ is uniformly continuous, then it extends to a unique uniformly continuous function $g$ on all of $\mathbb{R}$, and we have $f=g$ on $S$.

If $S$ contains a negative irrational number $x$, then $g(x) = f(x) = -1$. Let $y$ be any positive number in $S$. If $y$ is rational, we have $g(y) = f(y) = 0$. Then by the continuity of $g$, there would have to be some $z \in S$ with $f(z) = g(z) \in (-3/4, -1/4)$ which is impossible. If $y$ is irrational, we get a similar contradiction since $g(y) = f(y) = -2$. So $S$ does not contain any negative irrational. Similarly, $S$ does not contain any positive irrational.

So we must have $S \subset \mathbb{Q}$. But this is similarly impossible. The rationals in $S$ cannot all have the same denominator (in lowest terms), so let $x_1 = p_1/q_1, x_2 = p_2/q_2 \in S$, where $q_1 < q_2$. Then by the continuity of $g$ there must be some $y \in S$ with $f(y) = g(y) \in (\frac{1}{q_1}, \frac{1}{q_1+1})$, but $f$ never takes on any such value.

Consider the following modification of the Dirichlet "popcorn" function: $$f(x) = \begin{cases} 1/q, & \text{$x \in \mathbb{Q}$, $x=p/q$ in lowest terms} \\ -1, & x \notin \mathbb{Q},\, x < 0 \\ -2, & x \notin \mathbb{Q}, \, x > 0.\end{cases}$$ Since every real number can be approximated by rationals with arbitrarily large denominator, the closure of the graph of $f$ contains the $x$-axis, which is the uniformly continuous function $0$.

Let $S \subset \mathbb{R}$ be dense. If $f|_S$ is uniformly continuous, then it extends to a unique uniformly continuous function $g$ on all of $\mathbb{R}$, and we have $f=g$ on $S$.

If $S$ contains a negative irrational number $x$, then $g(x) = f(x) = -1$. Let $y$ be any positive number in $S$. If $y$ is rational, we have $g(y) = f(y) = 0$. Then by the continuity of $g$, there would have to be some $z \in S$ with $f(z) = g(z) \in (-3/4, -1/4)$ which is impossible. If $y$ is irrational, we get a similar contradiction since $g(y) = f(y) = -2$. So $S$ does not contain any negative irrational. Similarly, $S$ does not contain any positive irrational.

So we must have $S \subset \mathbb{Q}$. But this is similarly impossible. The rationals in $S$ cannot all have the same denominator (in lowest terms), so let $x_1 = p_1/q_1, x_2 = p_2/q_2 \in S$, where $q_1 < q_2$. Then by the continuity of $g$ there must be some $y \in S$ with $f(y) = g(y) \in (\frac{1}{q_1}, \frac{1}{q_1+1})$, but $f$ never takes on any such value.