Closed subset of a metric space admits open subsets

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I have the following question and I require some help.

Let $(X,d)$ be a metric space. Let $C$ be a closed subset of $X$, and let $x \in X$ be such that $x \notin C$. Show that there exist disjoint subsets $U, V$ of $X$ with $ C \subseteq U$ and $x \in V$.

I am also given the following hint:
Hint: you may consider the function $f: X \rightarrow \mathbb{R}$ defined by \begin{equation} f(y)=\inf\, \{d(y,c): c \in C\} \end{equation}
I have managed to show that the function in the hint is continuous, which I think will help to use it in combination with the fact that the inverse image of an open set is open, but I cannot exactly move forward from that.

Any solutions or suggestions will be helpful.
Thanks.

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Take $f^{-1}(\frac{f(y)}2,\infty)$ for $V$ ; and let $U=X\setminus {\overline{V}}$.

(You could also just take $V=B_y(\frac{f(y)}2)$, the open ball of radius $\frac{f(y)}2$ centered at $y$...)