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Let $f$ be a continuous function. Prove that the set $\{x\in\mathbb{R}^n\mid f(x)=0\}$ is not open.

I can see that this is a closed set... But how can I show that it's not open? Thanks.

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    If $f(x)=0$ for all $x\in\mathbb R^n$, then it is open. Similarly if $f(x)\neq0$ for all $x$. Did you include all of the details to this problem? – 2017-02-23
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    If the zero set is also open, then as it is both open and closed, it is a union of components of $\Bbb R^n$. In particular, if the domain of the function is all of $\Bbb R^n$ (which in particular is connected), openness of the zero set forces the function to be the zero function. – 2017-02-23
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    @Aweygan Sorry, I forgot. We assume that $f$ is non-constant and the set mentioned is not empty. – 2017-02-23
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    Then use the fact that $\mathbb R^n$ is connected (assuming you're working with the standard topology) to show that $\{x\in\mathbb{R}^n\mid f(x)=0\}$ is not open. – 2017-02-23

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As indicated in question, the set $Z=\{x\in\mathbb{R}^n\mid f(x)=0\} $ is closed. If in addition, $Z$ is open then because of connectedness of $\mathbb{R}^n$, we have either $Z=\varnothing$ or $Z=\mathbb{R}^n $ which implies $f$ is non-zero every where or constantly zero.