$f$ continuous, $f: X \to Y$, $Y$ compact Hausdorff. Is $f(X)$ compact? The question "Is $f(X)$ compact?" is something that occured to me when attempting the Munkres question.
I think $f(X)$ is compact. Let $ \{V_\alpha \}$ be an arbitrary open cover of $Y$ such that $ \bigcup_\alpha V_\alpha \supset Y \supset f(X).$ $\{V_\alpha \}$ is an open cover of $f(X)$.
Since $Y$ is compact, then there exists a finite subcover such that $ f(X) \subset Y \subset \bigcup_{i=1}^n V_{\alpha_i}$.
Lemma 26.1 of Munkres : Let $Y$ be a subspace of $X$. Then $Y$ is compact iff every covering of $Y$ by sets open in $X$ contains a finite subcollection covering $Y$.
So by the above lemma, $f(X)$ is compact as a subspace of $Y$.
I am not sure if it is right though, please kindly point out the mistakes. 
Thank you.
 A: Let $X$ be $\Bbb Z^+$ with the discrete topology, let $Y=[0,1]$ with the usual topology, and let $$f:X\to Y:n\mapsto\frac1n\;.$$ Then $f$ is continuous, $Y$ is compact Hausdorff, and $f[X]$ is not even closed in $Y$, let alone compact. A very simple open cover of $f[X]$ that has no finite subcover is 
$$\left\{\left(\frac1{n+1},\frac1{n-1}\right):n\ge 2\right\}\cup\left\{\left(\frac12,1\right]\right\}\;,$$
and it illustrates what’s wrong with your argument: it covers $f[X]$, but it *doesn’t cover $Y$. It covers only $(0,1]$.
A: Note that the continuity of $f$ is not enough to conclude that $f(X)$ is closed in $Y$. If $f(X)$ is closed then it is compact because a closed subset of a compact set is compact.
For example, fix $\{q_n\mid n\in\Bbb N\}$ an enumeration of $\Bbb Q\cap[0,1]$, then $n\mapsto q_n$ is continuous from the discrete space $\Bbb N$ into $[0,1]$ (every function from a discrete space is continuous), but $\Bbb Q\cap[0,1]$ is certainly not compact nor it is closed.
A: It suffices to find a non-compact subset $X$ of a compact Hausdorff space $Y$, because the inclusion map $i:X\hookrightarrow Y$ is always continuous. An easy one to consider is $X=(0,1)$ and $Y=[0,1]$. 
