While in general one cannot expect density of Lipschitz mappings (as pointed out by Nik Weaver), the following classical result is in the positive direction. You can find a proof in Heinonen's book Lectures on analysis on metric spaces:

Theorem. If $f:X\to\mathbb{R}$ is a bounded and uniformly continuous function on a metric space, then there is a sequence of Lipschitz continuous functions $f_i:X\to\mathbb{R}$, $i=1,2,3,\ldots$, such that $f_i\to f$ converges uniformly on $X$.

$X$ can be any metric space and compactness is not required. For that reason the argument based on the Stone-Weierstrass theorem mentioned in some comments cannot be applied here.

While in general one cannot expect density of Lipschitz mappings (as pointed out by Nik Weaver), the following classical result is in the positive direction. You can find a proof in Heinonen's book Lectures on analysis on metric spaces:

Theorem. If $f:X\to\mathbb{R}$ is a bounded and uniformly continuous function on a metric space, then there is a sequence of Lipschitz continuous functions $f_i:X\to\mathbb{R}$, $i=1,2,3,\dotsc$, such that $f_i\to f$ converges uniformly on $X$.

$X$ can be any metric space and compactness is not required. For that reason the argument based on the Stone–Weierstrass theorem mentioned in some comments cannot be applied here.

While in general one cannot expect density of Lipschitz mappings (as pointed out by Nik Weaver), the following classical result is in the positive direction. You can find a proof in Heinonen's book Lectures on analysis on metric spaces:

Theorem. If $f:X\to\mathbb{R}$ is a bounded and uniformly continuous function on a metric space, then there is a sequence of Lipschitz continuous functions $f_i:X\to\mathbb{R}$, $i=1,2,3,\ldots$, such that $f_i\to f$ converges uniformly on $X$.

$X$ can be any metric space and compactness is not required.

While in general one cannot expect density of Lipschitz mappings (as pointed out by Nik Weaver), the following classical result is in the positive direction. You can find a proof in Heinonen's book Lectures on analysis on metric spaces:

Theorem. If $f:X\to\mathbb{R}$ is a bounded and uniformly continuous function on a metric space, then there is a sequence of Lipschitz continuous functions $f_i:X\to\mathbb{R}$, $i=1,2,3,\ldots$, such that $f_i\to f$ converges uniformly on $X$.

$X$ can be any metric space and compactness is not required. For that reason the argument based on the Stone-Weierstrass theorem mentioned in some comments cannot be applied here.