Remainder function is a contraction I cannot solve the following problem:

Let $f:U\subset\mathbb{R}^m\rightarrow \mathbb{R}^n$ be a differentiable function in $a\in U$, such that $f'$ is continuous in $a$. Then for every $\epsilon >0$ exists $\delta>0$ such that
$$|r(v_1)-r(v_2)|\leq \epsilon |v_1-v_2|$$
for every $v_1,v_2\in B_{\delta}(0)$

Here the function $r$ comes from the definion of differentiability: $f:U\subset\mathbb{R}^m\rightarrow \mathbb{R}^n$ is differentiable in $a\in U$ if for every $v\in \mathbb{R}^m$, $a+v \in U$, exists a linear map $T:\mathbb{R}^m \rightarrow \mathbb{R}^n$ such that
$$f(a+v)=f(a)+Tv+r(v)  \hspace{1cm} \lim_{v \to 0} \frac{r(v)}{|v|}=0$$
Can someone help me?
 A: I will present a proof below which requires $f$ to be differentiable in a neighborhood of $a$ and $f'$ only need to be continuous at $a$. It will be divided in two parts. In Part 1, we reduce the problem to a local version, then in Part 2 we prove it.
Consider $\epsilon>0$ to be fixed.
Part 1 It suffices to prove that there exists $\delta>0$ satisfying the following:

For any $x\in B_\delta(0)$, there exists $\delta'>0$ such that $B_{\delta'}(x) \subset B_\delta(0)$ and
$$
\|r(x)-r(y)\|\leq \epsilon\|x-y\|,\ \forall y \in B_{\delta'}(x).
$$

Indeed, assume that the local result is proved and, in order to extend the estimate for any pair $x, y \in B_{\delta}(0)$, consider $S$ the closed line segment joining $x$ and $y$, that is,
$$
S=\{(1-\alpha)x + \alpha y : \alpha\in[0,1]\}.
$$
By the convexity of the ball, $S\subset B_{\delta}(0)$ and we can fix, for each $u\in S$, $\delta'_u>0$ satisfying the local result. Since $S$ is compact, we get $s_1, ..., s_n\in S$ such that the balls $B_{\delta'_{s_1}}(s_1), ..., B_{\delta'_{s_n}}(s_n)$ cover $S$. For any $s_i$, consider the corresponding $\alpha_i\in[0,1]$ such that
$$
s_n = (1-\alpha_n)x + \alpha_n y,
$$
and assume, without loss of generality, that $0=\alpha_1< \alpha_2 < ...< \alpha_n=1$. Now, using the connectedness of $S$, we can also assume that for each $i\in\{1,..., n-1\}$, there exists $\alpha_i<\beta_i< \alpha_{i+1}$ such that $$ t_i = (1-\beta_i)x+ \beta_i y\in B_{\delta'_{s_i}}(s_i)\cap B_{\delta'_{s_{i+1}}}(s_{i+1}).$$
Therefore,
\begin{align*}
\|r(x)-r(y)\| & = \left\|\sum_{i=1}^{n-1}(r(s_i)-r(t_i)) + (r(t_i)-r(s_{i+1}))\right\| \\
& \leq \sum_{i=1}^{n-1}\|r(s_i)-r(t_i)\| + \|r(t_i)-r(s_{i+1})\| \\
& \leq \epsilon\sum_{i=1}^{n-1}\|s_i-t_i\| + \|t_i-s_{i+1}\| \\
& = \epsilon\|x-y\|.
\end{align*}
Part 2 Now we prove the local result. Further, for each point $x\in U$, we denote $T_x= f'(x)$ and $r_x$ such that
$$
f(x+v) = f(x) + T_x(v) + r_x(v),\ \ \mbox{with }\ \lim_{v\to 0} \frac{r_x(v)}{v}=0.
$$
So what we want to prove is that there exists $\delta>0$ satisfying

For any $x\in B_\delta(0)$, there exists $\delta'>0$ such that $B_{\delta'}(x) \subset B_\delta(0)$ and
$$
\|r_a(x)-r_a(y)\|\leq \epsilon\|x-y\|,\ \forall y \in B_{\delta'}(x).
$$

Since $f'$ is continuous at $a$, fix $\delta>0$ such that
$$ \tag{1}\label{delta}
\|T_a - T_x\|\leq \frac{\epsilon}{2},\ \forall x\in B_\delta(a),
$$
where the norm considered here is the one for linear operators.
Fix $x\in B_\delta(0)$. Since $\lim_{h\to 0}\frac{r_{a+x}(h)}{\|h\|}=0$, fix $\delta'>0$ such that
$$\tag{2}\label{delta'}
\frac{\|r_{a+x}(h)\|}{\|h\|}\leq \frac{\epsilon}{2},\ \forall \|h\|< \delta'(0),
$$
and we may fix it in order to also satisfy $B_{\delta'}(x)\subset B_\delta(0)$.
Further, we finish the proof by showing that
$$
\|r_a(x)-r_a(y)\|\leq \epsilon\|x-y\|,\ \forall y \in B_{\delta'}(x).
$$
Fix $y \in B_{\delta'}(x)$. Then $y= x+h$, with $\|h\|<\delta'$. Evaluating $r_a(x)$ and $r_a(y)$, we get
\begin{align*}
r_a(x) - r_a(y) & = f(a+x)-f(a+y) - T_a(x-y) \\ & = f(a+x)-f((a+x)+h) + T_a(h),
\end{align*}
then
\begin{align*} \tag{3}\label{sub}
\frac{\|r_a(x) - r_a(y)\|}{\|x-y\|} & = \left\|\frac{f(a+x)-f((a+x)+h)}{\|h\|} + \frac{T_a(h)}{\|h\|}\right\|.
\end{align*}
On the other hand, note that
$$
r_{a+x}(h) = f((a+x)+h) - f(a+x) - T_{a+x}(h),
$$
so
$$
\frac{f(a+x) - f((a+x)+h)}{\|h\|} = \frac{- r_{a+x}(h) - T_{a+x}(h)}{\|h\|}.
$$
Substituting this on \eqref{sub}, we get
\begin{align*}
\frac{\|r_a(x) - r_a(y)\|}{\|x-y\|} & = \left\|\frac{- r_{a+x}(h) - T_{a+x}(h)}{\|h\|} + \frac{T_a(h)}{\|h\|}\right\| \\
& \leq \frac{\|r_{a+x}(h)\|}{\|h\|} + \frac{\|(T_{a+x}-T_a)(h)\|}{\|h\|},
\end{align*}
then, by \eqref{delta} and \eqref{delta'},
\begin{align*}
\frac{\|r_a(x) - r_a(y)\|}{\|x-y\|} & \leq \frac{\epsilon}{2}  + \frac{\|T_{a+x}-T_a\|\|h\|}{\|h\|} \leq \frac{\epsilon}{2}+\frac{\epsilon}{2}= \epsilon,
\end{align*}
and we are done.
