Proof of Mean value theorem in $R^{n}$ using line segment trick

For reference, my book is Mutlidimensional analysis. The page is 56. The proof of Mean value theorem in my book goes as follows. Firstly, consider a lemma(I'll state without proof).

Consider an open set $U \subset \mathbb{R}^{n}$ with any two fixedpoint $x, x' \in \mathbb{R}^{n}$ such that the line segment$$L(x', x) = \{x'+ t(x - x'): t \in [0, 1]\}$$ is contained within $U$. Define a differentiable function $f: U \rightarrow \mathbb{R}^{p}$,then for all $a \in \mathbb{R}^{p}$ there exists $\zeta(a)$ such that$$\langle f(x) - f(x'), a\rangle = \langle Df(\zeta(a))(x - x'), a\rangle$$Here,$Df(\zeta(a))$ is the total differentiation of $f$ at $\zeta(a)$.

Plugging in $a := f(x) - f(x')$, we clearly have$$\Vert f(x) - f(x')\Vert^{2} = \langle Df(\zeta(a))(x - x'), f(x) - f(x') \rangle \leq \Vert Df(\zeta(a))(x - x') \Vert \Vert f(x) - f(x') \Vert$$For $A \in Lin(\mathbb{R}^{n}, \mathbb{R}^{p}), h \in \mathbb{R}^{n}$, we do have$$\Vert Ah \Vert \leq \Vert A \Vert_{Eucl}\Vert h \Vert$$It then follows that$$\Vert f(x) - f(x') \Vert < \Vert Df(\zeta(a))(x - x')\Vert \leq \Vert Df(\zeta(a))\Vert_{Eucl}\Vert x - x' \Vert$$They actually conclude that$$\Vert f(x) - f(x') \Vert < \sup_{\zeta \in L(x, x')}\Vert Df(\zeta) \Vert_{Eucl}\Vert x - x' \Vert$$, which is absurd since we do not assume the set$$\{\Vert Df(\zeta)\Vert_{Eucl}\Vert x - x' \Vert: \zeta \in L(x, x')\}$$is bounded. Then, the author's statement of Mean value theorem.

Let $U$ be a convex, open subset of $\mathbb{R}^{n}$ and $f: U \rightarrow \mathbb{R}^{p}$ be a differentiable function. The function$Df$ exists as a consequence and assumed to be bounded in a sense that$$\forall h \in \mathbb{R}^{p}, \zeta \in \mathbb{R}^{n}: \VertDf(\zeta)h \Vert \leq k\Vert h \Vert$$ for some fixed $k \in \mathbb{R}$. Then, it holds that $f$ is Lipchitz continuous with $k$.Namely, $$\forall x, x' \in U: \Vert f(x) - f(x') \Vert < k\Vert x -> x'\Vert$$

My attempt:It exists$$\Vert A \Vert = sup\{\Vert Ah \Vert: \Vert h \Vert = 1\}$$We can then prove that$$\Vert A \Vert < \Vert A \Vert_{Eucl} < \sqrt{n}\Vert A \Vert$$Plugging in $A := Df(\zeta)$, we conclude that$$\Vert Df(\zeta) \Vert_{Eucl}\Vert x - x'\Vert < \sqrt{n}\Vert Df(\zeta) \Vert\Vert x - x' \Vert < \sqrt{n}k\Vert x - x' \Vert$$, which proves the constant $\sqrt{n}k$ but not $k$. How to approach from here?