Is the dot product on $\mathbf{R}^n$ ($n\ge 2$) Lipschitz?

I find that if I "reduce" the dimensions more, I can recover the equal sign in (1), and thus the 1D argument works again.

Let $x,u,v$ be three real numbers and $e_1=(1,0,\cdots,0)$ the unit vector in $\mathbf{R}^n$.

Now consider $$ |f(xe_1,ue_1)-f(xe_1,ve_1)|=|x||u-v|=|x|\cdot \|ue_1-ve_1\|_{\mathbf{R}^n}\tag{1} $$ and $$ \|(xe_1,ue_1)-(xe_1,ve_1)\|_{\mathbf{R}^n\times \mathbf{R}^n} =\|ue_1 -ve_1\|_{\mathbf{R}^n}\tag{2} $$

Comparing (1) and (2), one can see that $f$ cannot be Lipschitz, since otherwise, there exists $K>0$ such that $$ |x|\cdot \|ue_1-ve_1\|_{\mathbf{R}^n} =|f(xe_1,ue_1)-f(xe_1,ve_1)| \le K\|ue_1 -ve_1\|_{\mathbf{R}^n}\tag{3} $$ for all $x,u,v\in\mathbb{R}$, which is impossible.

If one looks at (3) further, one can fix $u=1$ and $v=0$ in the whole argument to simplify the contradiction to the estimate $$ |x|\le K,\quad x\in\mathbf{R}\;. $$