Proof of Uniqueness of the Derivative of a Real Function on an Interval

Suppose, for a contradiction, that $L\neq M$. Then $L-M\neq0$, so $0<|L-M|$ by Rules of Arithmetic and Order in an Ordered Field (x). Put $\varepsilon=|L-M|/(1+1)$. Then $0<\varepsilon$, by Rules of Arithmetic and Order in an Ordered Field (ix) applied to $0<|L-M|$, and $\varepsilon+\varepsilon=|L-M|$.

The choices are made in this order. First, the hypothesis for $N=L$, applied to this $\varepsilon$, gives a real $\delta_{L}>0$ such that every $t\in I$ with $0<|t-x|<\delta_{L}$ satisfies $|q(t)-L|<\varepsilon$, where $q(t)=(f(t)-f(x))/(t-x)$. Second, the hypothesis for $N=M$, applied to the same $\varepsilon$, gives a real $\delta_{M}>0$ such that every $t\in I$ with $0<|t-x|<\delta_{M}$ satisfies $|q(t)-M|<\varepsilon$. Third, let $\delta=\min\{\delta_{L},\delta_{M}\}$, a real number with $0<\delta$, $\delta\le\delta_{L}$ and $\delta\le\delta_{M}$. Fourth, $I$ is an interval containing two distinct points, so Basic Facts about Intervals of the Real Line (vii), applied to $I$, $x$ and $\delta$, gives $t\in I$ with $0<|t-x|<\delta$.

For this $t$ we have $|t-x|<\delta_{L}$ and $|t-x|<\delta_{M}$, so $|q(t)-L|<\varepsilon$ and $|q(t)-M|<\varepsilon$. Since $L-q(t)=-(q(t)-L)$, we have $|L-q(t)|=|q(t)-L|$ by Rules of Arithmetic and Order in an Ordered Field (x). As $L-M=(L-q(t))+(q(t)-M)$, the triangle inequality Rules of Arithmetic and Order in an Ordered Field (xi) gives

$$|L-M|\le|L-q(t)|+|q(t)-M| .$$

Adding the two strict inequalities, by Rules of Arithmetic and Order in an Ordered Field (iv) applied twice,

$$|L-q(t)|+|q(t)-M|<\varepsilon+|q(t)-M|<\varepsilon+\varepsilon=|L-M| .$$

Hence $|L-M|<|L-M|$, which is impossible, since $<$ is the strict relation of the order on $\mathbb{R}$. Therefore $L=M$, which is (i).