I really want to prove the statement in the title but I'm struggling with it. Here my current state:

Proof via contradiction. Let $\mathcal{H}$ be a RKHS with two reproducing kernels $k$ and $\hat{k}$ and let $x \in \mathcal{H}$. Then:

\begin{align} \|{k_x - \hat{k}_x}\|^2 &= \langle k_x - \hat{k}_x, k_x - \hat{k}_x \rangle \\ &= \langle k_x - \hat{k}_x , k_x \rangle - \langle k_x - \hat{k}_x , \hat{k}_x \rangle \\ &= \color{orange}{\langle k_x, k_x \rangle + \langle \hat{k}_x, \hat{k}_x \rangle} - \color{blue}{\langle \hat{k}_x, k_x \rangle - \langle k_x, \hat{k}_x \rangle}\\ &= ~... \\ &= \color{orange}{k(x,x) - \hat{k}(x,x)} - \color{blue}{k(x,x) + \hat{k}(x,x)} \\ &= 0. \end{align}

And this would be a contradiction since $\|x-y\| = 0 \Longleftrightarrow x = y$.

So the orange terms look fine but I don't know how to get the blue terms from the third to the fifth line. Please help.

Cheers. :-)