Suppose that given $h\in H$, there exists some $x\in [a,b]$ such that $h'(x)=0$.
Because $h$ is injective, we cannot have $h'\equiv 0$ on any open interval, so there is some open interval $(c,d)\subset [a,b]$ containing $x$ such that $|h'(t)|>0$ for all $t\in(c,d)$, $t\neq x$. Because $h'$ is continuous, $h'(t)$ has the same sign for all $t\in(c,x)$.
Because $h'$ is continuous and $h'(x)=0$, for any $\epsilon>0$ we can choose our interval $(c,d)$ containing $x$ to also have the property that $|h'(t)|<\frac{\epsilon}{4}$ for all $t\in (c,d)$.
For any $r,s$ with $c, we can create a non-zero $C^\infty$ function $w:[a,b]\to\mathbb{R}$ supported on $[r,s]\subset (c,d)$ such that
- $\|w\|<\epsilon$
- $|w'(z)|>\frac{\epsilon}{2}$ for some $z\in (r,x)$
- $w'$ has the opposite sign of $h'$ on $(r,x)$
Thus $\|(h+w)-h\|=\|w\|<\epsilon,$ and $h+w\in C^1([a,b],\mathbb{R})$, but the sign of $(h+w)'$ changes between $r$, where $(h+w)'(r)=h'(r)+w'(r)=h'(r)+0=h'(r)$ has the same sign as $h'$, and $z$, where $(h+w)'(z)=h'(z)+w'(z)$ has the opposite sign of $h'$ because $w'$ has the opposite sign of $h'$ on $(r,x)$ and $|w'(z)|>\frac{\epsilon}{2}>\frac{\epsilon}{4}>|h'(z)|$.
Thus $h+w$ is not injective, and thus $h+w\notin H$. Because, for any $\epsilon>0$, we can find some $h+w$ such that $\|(h+w)-h\|<\epsilon$ and $h+w\notin H$, the set $H$ is not closed in $C^1([a,b],\mathbb{R})$.