Fact 1 (Goldowsky-Tonelli): Let $F:(a, b) \to \mathbb{R}$ be continuous and have finite derivative everywhere. Suppose $F' \geq 0$ almost everywhere. Then $F$ is monotonically increasing.
For a proof of this, see Saks, Theory of the integral, Chapter 6, page 206.
Suppose $X = \{x \in [a, b]: -1 < f'(x) < 1\}$ has zero measure. Let $Y = \{x \in [a, b]: f'(x) \leq -1\}$ and $Z = \{x \in [a, b]: f'(x) \geq 1\}$
Claim 1: Every point of $X$ is a limit point of $Y$ and a limit point of $Z$.
Proof: Suppose for example $x \in X$ is not a limit point of $Y$ - the other case is similar. Let $I$ be an open interval around $x$ disjoint with $Y$. Then at almost every $y \in I$, $f'(y) \geq 1$. Using Fact 1, it follows that the function $y \mapsto f(y) - y$ is monotonically increasing on $I$ and hence for every $y \in I$, $f'(y) \geq 1$ which is impossible as $x \in Y \cap X$.
Claim 2: The set of points of continuity of $f' \upharpoonright \overline{X}$ is dense in $\overline{X}$ (the closure of $X$).
Proof: Well known (using Baire category theorem).
Now let $I$ be any open interval around $x \in X$. Then the supremum of $f' \upharpoonright I$ is at least $1$ and its infimum is at most $-1$ by Claim 1. By Darboux theorem, it follows that $f' \upharpoonright I$ and hence also $f' \upharpoonright (I \cap X)$ takes every value in $(-1, 1)$. So $f'$ is everywhere discontinuous on $\overline{X}$ which contradicts Claim 2.
Best Answer
Proof of the implicit function theorem in several variables calculus requires the contraction mapping theorem, so is probably not suitable for your audience. You need to use an iterative method and take a limit. You can look for a complete proof in Spivak, Calculus on Manifolds.
If you just replace one of the coordinate functions by $f$, you get a chart, but the proof that it is a chart requires the implicit function theorem.