Here is my idea:
Suppose $f : \mathbb S^1 \to \mathbb S^1$ is continuous and injective.
If $f$ is not surjective, its image is homeomorphic to a subset of $\mathbb S^1 \setminus \{ c\} \cong \mathbb R$ (by considering the projection).
Then, the restriction of $f$'s codomain $g : \mathbb S^1 \to f(\mathbb S^1)$ is injective and surjective, so it is bijective.
Also, since $\mathbb S^1$ is compact and connected, its image under the continuous map is also compact and connected, so $f(\mathbb S^1)$ is homeomorphic to some closed interval $[a, b]$, which is simply connected.
Noting $f(\mathbb S^1)$ is Hausdorff and $\mathbb S^1$ is compact, by the closed map lemma, it follows $g$ is a homeomorphism. This contradicts that $\mathbb S^1$ is not simply connected, since $[a, b]$ is simply connected. Thus, $f$ must be surjective.
Does it work?
Best Answer
This is basically the same argument, but you can be a little bit shorter by noticing that any non-surjective continuous map $f:X \rightarrow \mathbb{S}^1$, where $X$ is a metric space, can be lifted to a map $\tilde{f} : X \rightarrow \mathbb{R}$. Of course, if $X=\mathbb{S}^1$, then $\tilde{f}$ cannot be injective, so $f$ is not injective.