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I want to prove that $\text C(\Omega)$ do not have Heine-Borel Property. Here $\Omega$ is a countable union of compact sets $\ K_n$ where $n\in \mathbb N$ and $K_{n}^\circ \subset K_{n+1}$. I don't have any idea how to proceed it, can anyone please help me in it.

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    What norm or metric are you giving $C(\Omega)$? Because if $\Omega$ is not compact then the sup norm is not well defined on $C(\Omega)$. – 2017-02-25
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    It should be $K_n^o\subset K_{n+1}$, rather that $K_n^o\in K_{n+1}$. – 2017-02-25
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    metric on $C(\Omega)$ is given in "Functional Analysis-Rudin" Page No. 33 – 2017-02-25
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    @bunny you should add the definition to your question. Not everybody has that book and not everybody want to look into it. – 2017-02-25
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    What does $K_n^\circ$ mean? – 2017-02-25

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I guess you mean the the locally convex topology $\tau$ of uniform convergence on each $K_n$. This can be given by a metric but also (from the locally convex point of view: more naturally) by the family of seminorms $$\|f\|_n=\sup\lbrace|f(x)|:x\in K_n\rbrace.$$ Without further assumption on $\Omega$ it may happen that this locally convex space does have the Heine-Borel property (usually, spaces where all bounded sets are relatively compact are called Montel spaces). This is the case for $\Omega=\mathbb N$ (with the discrete topology, of course). Then $C(\Omega)$ is the space $\mathbb R^\mathbb N$ of all scalar sequences and $\tau$ is the product topology which has the Heine-Borel property by Tychonov's theorem.

$C(\Omega)$ fails to be a Montel space whenever one of the subspaces $X_n=\lbrace f\in C(\Omega): f=0 \text{ on } \Omega\setminus K_n\rbrace$ is infinite dimensional because $X_n$ with the relative topology induced by $\tau$ is a Banach space and infinite dimensional Banach spaces are never Montel.