I read that $x \mapsto \sqrt{\mid x \mid}$ is a quasi-convex function. However, the sub-level sets still look like two curves each with its opening facing downwards. 
Why is it called quasi-convex?
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ViktorStein
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Surya Narayanan
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Have a look here – Ninad Munshi Jun 27 '20 at 04:57
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The $\alpha$ sub-level sets are not curves, but rather subsets of the domain (so in this case, parts of the $x$-axis, or the real line).
In this case, you can show that for any $\alpha \ge 0$, the $\alpha$ sub-level set of the function $f:\Bbb{R}\to \Bbb{R}, f(x)=\sqrt{|x|}$ is
$$ \{ x\in \Bbb{R}: f(x)\le \alpha\} = \left[-\alpha^2,\alpha^2\right].$$
Since this is a convex subset of $\Bbb{R}$ (the domain of $f$), $f$ is quasiconvex.
Minus One-Twelfth
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A function is quasi-convex if its lower contour sets, $L(\alpha)=\{x:f(x) \le \alpha \}$, are convex for all $\alpha \in \mathbb{R}$. If you consider $|x| \le \alpha^2$, it's the set of points for which $ -\alpha^2 \le x \le \alpha^2$. That is indeed a convex set. Therefore $f$ is quasi-convex.
Quasi-convexity in $\mathbb{R}$ is a strange beast though. Any strictly increasing or decreasing function from $\mathbb{R}$ to $\mathbb{R}$ is both quasi-convex and quasi-convex, for example. It's a much more useful definition for minimization in $\mathbb{R}^N$.
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Yes, quasi-concave functions have convex upper contour sets, $U(\alpha) = {x:f(x) \ge \alpha}$. So a monotone function $f : \mathbb{R} \rightarrow \mathbb{R}$ is both quasi-concave and quasi-convex, since both its upper and lower contour sets are convex. Your intuition from concave and convex functions can be very misleading. – Jun 27 '20 at 14:03