Let $\mathbb{R}_u$ be $\mathbb{R}$ equipped with the usual topology and $\mathbb{R}_K$ be $\mathbb{R}$ equipped with the $K$-topology.
Is there a function $f : \mathbb{R} \to \mathbb{R}$ which is continuous when the domain is $\mathbb{R}_K$ but not when it is $\mathbb{R}_u$? (The codomain is taken with usual topology in both cases.)
No. More generally, for any regular space $X$, the continuous maps $\mathbb{R}_K\to X$ and $\mathbb{R}_u\to X$ are the same. To prove this, suppose $f:\mathbb{R}_K\to X$ is continuous and let $U\subseteq X$ be open. Since $f^{-1}(U)$ is open in $\mathbb{R}_K$, it contains an open interval around all its points except for possibly $0$, so we just have to show it contains an open interval around $0$ if $0\in f^{-1}(U)$. Now note that by regularity, there is a closed neighborhood $C$ of $f(0)$ contained in $U$. Thus $f^{-1}(C)\subseteq f^{-1}(U)$ is a closed neighborhood of $0$ in $\mathbb{R}_K$. But the closure of any neighborhood of $0$ in $\mathbb{R}_K$ contains an open interval around $0$ (since $1/n$ will be in the closure for all sufficiently large $n$), and so $f^{-1}(U)$ does contain an open interval around $0$.