Proof of Restriction Stability of Continuity and of the Derivative

Claim 1. Let $\varepsilon$ be a real number with $\varepsilon>0$. Since $f$ is continuous at $x$ relative to $A$, there exists a real $\delta>0$ such that every $y\in A$ with $d_X(x,y)<\delta$ satisfies $d_Y(f(y),f(x))<\varepsilon$. Let $y\in B$ satisfy $d_X(x,y)<\delta$. Since $B\subseteq A$ we have $y\in A$, and therefore

As $\varepsilon>0$ was arbitrary, $f|_B$ is continuous at $x$ relative to $B$.

For the consequence, suppose $f$ is continuous on $A$, and let $x\in B$ be arbitrary. Then $x\in A$, so $f$ is continuous at $x$ relative to $A$, and by the first part $f|_B$ is continuous at $x$ relative to $B$. Since this holds for every $x\in B$, the restriction $f|_B$ is continuous on $B$.

Claim 2. Since $x_0$ is an interior point of $J$, there exist $u,v\in J$ with $u<x_0<v$. As $J\subseteq I$ we have $u,v\in I$, so $x_0$ is an interior point of $I$.

Suppose $f$ is differentiable at $x_0$, write $L=f'(x_0)$, and let $|\cdot|$ denote the absolute value on $\mathbb{R}$. Let $\varepsilon>0$ be real. By the definition of the derivative there exists a real $\delta>0$ such that every real $h$ with $0<|h|<\delta$ and $x_0+h\in I$ satisfies

Now let $h$ be real with $0<|h|<\delta$ and $x_0+h\in J$. Then $x_0+h\in I$, and $f|_J(x_0+h)=f(x_0+h)$ and $f|_J(x_0)=f(x_0)$, so

As $\varepsilon>0$ was arbitrary, and $x_0$ is an interior point of the interval $J$, the restriction $f|_J$ is differentiable at $x_0$ with $(f|_J)'(x_0)=L=f'(x_0)$. $\blacksquare$