Proof of Uniqueness of the Partial Derivative on a Euclidean Open Set

Let $|\cdot|$ be the absolute value on $\mathbb{R}$. Suppose $L\ne L'$; then $L-L'\ne0$, so $|L-L'|>0$ by claim 1 of Properties of the Absolute Value in an Ordered Field, and $\varepsilon=|L-L'|/2$ is a positive real by claim 8 of Elementary Order Arithmetic in an Ordered Field.

By Partial Derivative on a Euclidean Open Set, applied once for the value $L$ and once for the value $L'$, there are reals $\delta>0$ and $\delta'>0$ such that every $h\in\mathbb{R}$ with $0<|h|<\delta$ (respectively $0<|h|<\delta'$) satisfies $(a_1,\dots,a_{i-1},a_i+h,a_{i+1},\dots,a_n)\in U$ together with the difference-quotient inequality for $L$ with $\varepsilon$ (respectively for $L'$ with $\varepsilon$).

Let $\eta$ be the lesser of $\delta$ and $\delta'$ (claim 9 of Elementary Order Arithmetic in an Ordered Field), positive since $\eta$ is $\delta$ or $\delta'$, and put $h=\eta/2$, so that $0<h<\eta$ by claim 8 there and $|h|=h$ since $0\le h$ (absolute value); hence $0<|h|<\delta$ and $0<|h|<\delta'$ by claim 2 of Elementary Order Arithmetic in an Ordered Field. Writing

$$q=\frac{f(a_1,\dots,a_{i-1},a_i+h,a_{i+1},\dots,a_n)-f(a)}{h},$$

both $|q-L|<\varepsilon$ and $|q-L'|<\varepsilon$ hold, and $|L-q|=|q-L|$ by claim 2 of Properties of the Absolute Value in an Ordered Field, so by the triangle inequality for the absolute value

$$|L-L'|\le|L-q|+|q-L'|<\varepsilon+\varepsilon=|L-L'| ,$$

a contradiction. Hence $L=L'$. $\blacksquare$