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Let $(X,\rho)$ be a compact metriv space, and let $(Y,d)$ be a separable metric space. Then I claim that one can endow $X$ with a compatible metric $d$ such that every continuous $f:X\rightarrow Y$ can be uniformly approximated by a Lipschitz function.

Let $(f_n)_{n\geq 0}$ be a sequence of continuous functions such that every continuous $f:X\rightarrow Y$ can be uniformly approximated by some $f_n$. Let $(\alpha_n)_{n\geq 0}$ be a sequence of positive real numbers such that $\alpha_n\cdot\text{Diam}(f_n[X])\rightarrow 0$. Then define a metric $d$ on $X$ by setting $$d(x,y)=\rho(x,y)+\sup_n\alpha_n\cdot d(f_n(x),f_n(y)).$$ Then the metric $d$ is compatible with the original topology on $X$. Furthermore, for each $n$, we have $d(f_n(x),f_n(y))\leq\alpha_n^{-1}\cdot d(x,y)$, so each $f_n$ is Lipschitz.

Let $(X,\rho)$ be a compact metric space, and let $(Y,d)$ be a separable metric space. Then I claim that one can endow $X$ with a compatible metric $d$ such that every continuous $f:X\rightarrow Y$ can be uniformly approximated by a Lipschitz function.

Let $(f_n)_{n\geq 0}$ be a sequence of continuous functions such that every continuous $f:X\rightarrow Y$ can be uniformly approximated by some $f_n$. Let $(\alpha_n)_{n\geq 0}$ be a sequence of positive real numbers such that $\alpha_n\cdot\operatorname{Diam}(f_n[X])\rightarrow 0$. Then define a metric $d$ on $X$ by setting $$d(x,y)=\rho(x,y)+\sup_n\alpha_n\cdot d(f_n(x),f_n(y)).$$ Then the metric $d$ is compatible with the original topology on $X$. Furthermore, for each $n$, we have $d(f_n(x),f_n(y))\leq\alpha_n^{-1}\cdot d(x,y)$, so each $f_n$ is Lipschitz.

Let $(X,\rho)$ be a compact metriv space, and let $(Y,d)$ be a separable metric space. Then I claim that one can endow $X$ with a compatible metric $d$ such that every continuous $f:X\rightarrow Y$ can be uniformly approximated by a Lipschitz function.

Let $(f_n)_{n\geq 0}$ be a sequence of continuous functions such that every continuous $f:X\rightarrow Y$ can be uniformly approximated by some $f_n$. Let $(\alpha_n)_{n\geq 0}$ be a sequence of positive real numbers such that $\alpha_n\cdot\text{Diam}(f_n[X])\rightarrow 0$. Then define a metric $d$ on $X$ by setting $d(x,y)=\rho(x,y)+\sup_n\alpha_n\cdot d(f_n(x),f_n(y)).$ Then the metric $d$ is compatible with the original topology on $X$. Furthermore, for each $n$, we have $d(f_n(x),f_n(y))\leq\alpha_n^{-1}\cdot d(x,y)$, so each $f_n$ is Lipschitz.

Let $(X,\rho)$ be a compact metriv space, and let $(Y,d)$ be a separable metric space. Then I claim that one can endow $X$ with a compatible metric $d$ such that every continuous $f:X\rightarrow Y$ can be uniformly approximated by a Lipschitz function.

Let $(f_n)_{n\geq 0}$ be a sequence of continuous functions such that every continuous $f:X\rightarrow Y$ can be uniformly approximated by some $f_n$. Let $(\alpha_n)_{n\geq 0}$ be a sequence of positive real numbers such that $\alpha_n\cdot\text{Diam}(f_n[X])\rightarrow 0$. Then define a metric $d$ on $X$ by setting $$d(x,y)=\rho(x,y)+\sup_n\alpha_n\cdot d(f_n(x),f_n(y)).$$ Then the metric $d$ is compatible with the original topology on $X$. Furthermore, for each $n$, we have $d(f_n(x),f_n(y))\leq\alpha_n^{-1}\cdot d(x,y)$, so each $f_n$ is Lipschitz.

Let $X$ be a compact metrizable space, and let $(Y,d)$ be a separable metric space. Then I claim that one can endow $X$ with a compatible metric $d$ such that every continuous $f:X\rightarrow Y$ can be uniformly approximated by a Lipschitz function.

Let $(f_n)_{n\geq 0}$ be a sequence of continuous functions such that every continuous $f:X\rightarrow Y$ can be uniformly approximated by some $f_n$. Let $(\alpha_n)_{n\geq 0}$ be a sequence of positive real numbers such that $\alpha_n\cdot\text{Diam}(f_n[X])\rightarrow 0$. Then define a metric $d$ on $X$ by setting $d(x,y)=\sup_n\alpha_n\cdot d(f_n(x),f_n(y)).$ Then the metric $d$ is compatible with the original topology on $X$. Furthermore, for each $n$, we have $d(f_n(x),f_n(y))\leq\alpha_n^{-1}\cdot d(x,y)$, so each $f_n$ is Lipschitz.

Let $(X,\rho)$ be a compact metriv space, and let $(Y,d)$ be a separable metric space. Then I claim that one can endow $X$ with a compatible metric $d$ such that every continuous $f:X\rightarrow Y$ can be uniformly approximated by a Lipschitz function.

Let $(f_n)_{n\geq 0}$ be a sequence of continuous functions such that every continuous $f:X\rightarrow Y$ can be uniformly approximated by some $f_n$. Let $(\alpha_n)_{n\geq 0}$ be a sequence of positive real numbers such that $\alpha_n\cdot\text{Diam}(f_n[X])\rightarrow 0$. Then define a metric $d$ on $X$ by setting $d(x,y)=\rho(x,y)+\sup_n\alpha_n\cdot d(f_n(x),f_n(y)).$ Then the metric $d$ is compatible with the original topology on $X$. Furthermore, for each $n$, we have $d(f_n(x),f_n(y))\leq\alpha_n^{-1}\cdot d(x,y)$, so each $f_n$ is Lipschitz.