Assume that $h_n \to h$ uniformly on the domain $D$ of $\{h_n\}$, where
$h_n(x) = \begin{cases} \frac{x}{n} \biggl(1 + \frac{1}{n} \biggr), & \text{if } x \text{ is irrational} \\ a + \frac{a}{n} + \frac{a}{b} \biggl(1 + \frac{1}{n} \biggr) \biggl(\frac{1}{n} \biggr), & \text{if } x \text{ is rational, say } x = a/b, b > 0 \end{cases}$
and $h(x) = \begin{cases} 0, & \text{if }x\text{ is irrational} \\ a, & \text{if }x\text { is rational, say }x = a/b, b > 0. \end{cases}$
If there is a positive rational in $D$ we can find an interval $A = [s,t]$ contained in $D$ with $s$ rational because $D$ is a bounded interval. Then given $\epsilon = 1$ there exists a positive integer $M$ such that $n \ge M$ implies that
\begin{align} \biggl|h_n\biggl(\frac{a}{b}\biggr) - h\biggl(\frac{a}{b}\biggr) \biggr| &= \ \biggl| \frac{a}{n} + \frac{a}{b} \biggl( \frac{1}{n} + \frac{1}{n^2} \biggr) \biggr| &< 1 \end{align}
whenever $n \ge$ M and $a/b \in A$ where $a$ and $b$ are positive integers. But then
\begin{align} \frac{a}{M} &< \frac{a}{M} + \frac{a}{b} \biggl( \frac{1}{M} + \frac{1}{M^2} \biggr) \ &= \biggl| \frac{a}{M} + \frac{a}{b} \biggl( \frac{1}{M} + \frac{1}{M^2} \biggr) \biggr| \ &< 1. \end{align}
Writing $s = c/d$ where $s$ is the left endpoint of $A$ and $c$ and $d$ are positive integers, we can choose a positive integer $k$ large enough so that both $(ck + 1)/d \in A$ and $ck + 1 > M$.
This contradicts $(ck + 1)/M < 1$. Therefore the assumption that $h_n \to h$ uniformly is false. A similar argument holds if no positive rationals are in $D$