Limit of sequence of infimums of a continuous function over converging sequence of sets equals infimum of the function over the limit of the set.

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Let $X$ be any topological space. Let $C_n$ be monotone decreasing sequence of compact sets in $X$ that converges to a non-empty compact set $C$. That is $\bigcap_{n\geq 0}C_n = C$ and $C_{n+1}\subseteq C_n$ for all $n$. Suppose $f:X\rightarrow \mathbb{R}$ be a continuous function. Then, is it true that $\lim\limits_{n\to \infty}(\inf\limits_{x\in C_n} f(x)) =\inf\limits_{x\in C} f(x) $?

My approach, one side of the inequality is straight forward. That is $\lim\limits_{n\to \infty}(\inf\limits_{x\in C_n} f(x)) \leq\inf\limits_{x\in C} f(x) $, because $C_n$ contains $C$ for all $n$. The other side inequality is not getting clear. Please help me.

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Answer to the question before compactness of $C_n$'s was added to the hypothesis. Let $X=\mathbb R, C_n=\{0\} \cup [n, \infty),C=\{0\}$. Then $f(x)=e^{-x}$ provides a counterexample since the infimum on $C_n$ is $0$ for each $n$ whereas in infimum on $C$ is $1$.