If $f$ is a loop in $\mathbb{S}^{n}$, then is $f^{-1}(\{x\})$ a compact set in $[0, 1]$? If $f$ is a loop in $\mathbb{S}^{n}$, then is $f^{-1}(\{x\})$, $x \in \mathbb{S}^{n}$, a compact set in $[0, 1]$?

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Since $\mathbb S^n$ is Hausdorff, singletons are closed. 
The loop $f : [0, 1] \to \mathbb S^n$ is a continuous map, so the inverse image of closed sets is closed, in particular $f^{-1}(\{x\})$ is closed.
But $[0, 1]$ is compact and closed subsets of compact sets are compact.
Conclude that $f^{-1}(\{x\})$ is compact.
