Proof of Euclidean Space is Open in Itself, and $C^k$ Maps are Continuous

Proof of claim 1. Let $x=(x_{1},\dots,x_{n})\in\mathbb{R}^{n}$ and take $r=1$, which satisfies $0<r$ by claim 6 of Elementary Order Arithmetic in an Ordered Field. Every point $y=(y_{1},\dots,y_{n})\in\mathbb{R}^{n}$ satisfying $\sum_{i=1}^{n}(y_{i}-x_{i})^{2}<r^{2}$ belongs to $\mathbb{R}^{n}$, since by hypothesis it is a point of $\mathbb{R}^{n}$. Thus the condition of Open Subset of Euclidean Space is met at every point of $\mathbb{R}^{n}$, and $\mathbb{R}^{n}$ is open in $\mathbb{R}^{n}$.

Proof of claim 2. By claim 6 of Arithmetic of Addition on the Natural Numbers, either $k=1$ or there is a natural number $j$ with $k=S(j)$, where $S$ is the successor map of Natural Numbers; and $S(j)=j+1$ by claim 1 of Arithmetic of Addition on the Natural Numbers. If $k=1$ there is nothing to prove. Otherwise $k=j+1$, and clause 2 of C^k Maps on a Euclidean Open Set says that a map is of class $C^{j+1}$ on $U$ precisely when it is of class $C^{1}$ on $U$ and its partial derivatives satisfy a further condition; in particular $F$ is of class $C^{1}$ on $U$.

Proof of claim 3. By claim 2, $F$ is of class $C^{1}$ on $U$. Clause 1 of C^k Maps on a Euclidean Open Set then gives, for every $j$ with $1\le j\le m$, that the coordinate function $F_{j}$ is continuous at every point of $U$ in the Euclidean sense.

Fix such a $j$ and a point $a\in U$. Claim 1 of Euclidean Continuity Agrees with Metric Continuity for Real-Valued Functions, applied with $E=U$, with $f=F_{j}$ and with this $a$, states that continuity of $F_{j}$ at $a$ in the Euclidean sense holds if and only if $F_{j}$ is continuous at $a$ relative to $U$ as a map from $U$ into $(\mathbb{R},d_{\mathbb{R}})$. The former holds, hence so does the latter; and $a\in U$ was arbitrary.

Finally, suppose $F$ is smooth on $U$. By Smooth Map on a Euclidean Open Set, $F$ is then of class $C^{k}$ on $U$ for every natural number $k$; taking $k=1$, the two preceding paragraphs apply and give the same conclusions. $\blacksquare$