$E$ is a normed vector space and $\Omega \subset E$ is open.
Let $f: \Omega \to \Bbb R$ be Fréchet differentiable on $\Omega$.
Let $Df$ be the derivative map of $f$; $Df: \Omega \to \mathcal L(E, \Bbb R)$, $x \mapsto Df(x)$, where $Df(x)$ is the differential map of $f$ at $x$; i.e., the linear map such that:
$$\lim_{h \to 0} \frac1{\|h\|} | f(x+h) - f(x) - Df(x) h | = 0$$
Suppose that $Df$ is continuous at $x_0$.
It's required to prove that for all $\epsilon > 0$, there exists $\delta > 0$, such that for all $s$ and $t$ in $E$, $\| s \| < \delta$ and $\| t \| < \delta$ imply that:
$$|f(x_0 + s) - f(x_0 + t) - Df(x_0) (s-t) | < \epsilon \| s - t \|$$
Any suggestions?
WLOG, $x_0 = 0.$ Let $\epsilon>0.$ Choose $\delta > 0$ such that $x\in B(0, \delta) \implies \|Df(x) - Df(0)\| < \epsilon.$
Fix $s,t\in B(0,\delta).$ Note that the map $l(u) = t + u(s-t)$ is a differentiable map from $\mathbb R$ to $E,$ with $Dl(u)[v] = v(s-t).$ So if we define $g = f\circ l,$ we see $g:\mathbb [0,1] \to \mathbb R$ is the composition of differentiable maps, hence is differentiable. By the mean value theorem
$$f(s) - f(t)=g(1) - g(0) = g'(c) = Dg(c)[1].$$
Now the chain rule says
$$Dg(c)[1] = (Df(l(c))\circ Dl(c))[1] =(Df(l(c))[Dl(c))[1]] = (Df(l(c))[s-t].$$
Rewrite the last term as
$$Df(0)[s-t] + (Df(l(c))-Df(0))[s-t].$$
Because $l(c) \in B(0,\delta),$ we get
$$|\,(Df(l(c))-Df(0))[s-t]\,| \le \|(Df(l(c))-Df(0))\|\ |s-t| < \epsilon|s-t|$$ as desired.