Do we need the surjectivity of a continuous $f: (X, \tau) \to (Y, \rho)$ to get $(X, \tau)$ compact, implies $(Y, \rho)$ compact?

Question

I found the following proposition in Morris' "Topology without tears":

Proposition: Let $f: (X, \tau) \to (Y, \rho)$ be a continuous surjective map. Thus, if $(X, \tau)$ is compact, then $(Y, \rho)$ is compact.

In the proof, the author starts from an open covering $\cup_{i \in I} O_i \supseteq Y$ and the compactness of $(X, \tau)$, to get a finite subcover of $(X, \tau)$, which is $\cup_{i \in I} f^{-1} (O_i)$. Then, he proceed with

\begin{align} Y & = f(X)\\ & \subseteq f \bigg( \bigcup_{k=1}^n f^{-1} (O_{i_k}) \bigg)\\ & = \bigcup_{k=1}^n f(f^{-1} (O_{i_k}))\\ & = \bigcup_{k=1}^n O_{i_k}. \end{align}

This last step should come from the fact that $f ( f^{-1} (x)) = x$ thanks to the surjectivity of $f$.

Question: Do we really need to assume the surjectivity of $f$ to establish the result?

To me it seems that it is used just for some pedagogical reason to make more explicit the fact that we end up with a finite subcovering of $Y$, thanks to the surjectivity of the function that allows the final step of the chain of equalities. Am I correct?

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Edit: As a matter of fact, I checked Munkres' book on topology, and he does not assume surjectivity to get the same result. Still, I actually buy Clayton's answer below.

So, now I am really puzzled. What is going on here?

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As always any feedback is most welcome.
Thank you for your time.

Answer 1

You need surjectivity otherwise an example like $f:[0,1]\to\Bbb R$ with $f(x)=x$ would show $\Bbb R$ is compact.