[proofplan]
We restrict the homeomorphism $f:X\to Y$ to the subspace $K\subset X$ and observe that this restriction is a homeomorphism from $K$ onto $f(K)$. Compactness is then proved directly from the open-cover definition: an [open cover](/page/Open%20Cover) of $f(K)$ pulls back to an open cover of $K$, and a finite subcover pushes forward to a finite subcover of $f(K)$. The reverse implication is the same argument applied to the inverse homeomorphism $f^{-1}:Y\to X$.
[/proofplan]
[step:Restrict the homeomorphism to the subset $K$]
Define the restricted map
\begin{align*}
g:K\to f(K), \quad g(x)=f(x).
\end{align*}
The map $g$ is bijective because $f:X\to Y$ is bijective. Its inverse is the restricted map
\begin{align*}
h:f(K)\to K, \quad h(y)=f^{-1}(y).
\end{align*}
Indeed, if $y\in f(K)$, then $y=f(x)$ for some $x\in K$, so $h(y)=x\in K$.
We verify continuity of $g$ in the subspace topologies. Let $U\subset f(K)$ be open in the [subspace topology](/page/Subspace%20Topology) inherited from $Y$. By definition of the subspace topology, there exists $V\in\tau_Y$ such that
\begin{align*}
U=V\cap f(K).
\end{align*}
Since $f:X\to Y$ is continuous, $f^{-1}(V)\in\tau_X$. Therefore
\begin{align*}
g^{-1}(U)=K\cap f^{-1}(V),
\end{align*}
which is open in $K$ with its subspace topology. Thus $g$ is continuous.
The same argument applied to $f^{-1}:Y\to X$ shows that $h:f(K)\to K$ is continuous. Hence $g:K\to f(K)$ is a homeomorphism of subspaces.
[/step]
[step:Pull back an open cover of $f(K)$ and extract a finite subcover of $K$]
Assume that $K$ is compact in the subspace topology inherited from $X$. Let $A$ be an index set, and let $(U_\alpha)_{\alpha\in A}$ be an open cover of $f(K)$ in the subspace topology inherited from $Y$. Thus each $U_\alpha\subset f(K)$ is open in $f(K)$, and
\begin{align*}
f(K)\subset \bigcup_{\alpha\in A}U_\alpha.
\end{align*}
For each $\alpha\in A$, the set $g^{-1}(U_\alpha)$ is open in $K$ because $g:K\to f(K)$ is continuous. The family $(g^{-1}(U_\alpha))_{\alpha\in A}$ covers $K$: if $x\in K$, then $g(x)=f(x)\in f(K)$, so there exists $\alpha\in A$ with $g(x)\in U_\alpha$, and hence $x\in g^{-1}(U_\alpha)$. Since $K$ is compact, there exist finitely many indices $\alpha_1,\dots,\alpha_m\in A$ such that
\begin{align*}
K\subset \bigcup_{j=1}^{m}g^{-1}(U_{\alpha_j}).
\end{align*}
[guided]
We want to prove compactness of $f(K)$ using the open-cover definition. So we start with an arbitrary open cover of $f(K)$ and show that it has a finite subcover.
Let $A$ be an index set, and let $(U_\alpha)_{\alpha\in A}$ be an open cover of $f(K)$ in the subspace topology from $Y$. This means that every $U_\alpha$ is open as a subset of the [topological space](/page/Topological%20Space) $f(K)$, and every point of $f(K)$ lies in at least one of these sets:
\begin{align*}
f(K)\subset \bigcup_{\alpha\in A}U_\alpha.
\end{align*}
The compactness assumption is about $K$, not about $f(K)$, so we transport the cover back along the restricted map $g:K\to f(K)$. For each $\alpha\in A$, define the subset $g^{-1}(U_\alpha)\subset K$. Because $g$ is continuous between the subspaces $K$ and $f(K)$, each $g^{-1}(U_\alpha)$ is open in $K$.
We now check that these inverse images cover $K$. Let $x\in K$. Then $g(x)=f(x)$ is an element of $f(K)$. Since the family $(U_\alpha)_{\alpha\in A}$ covers $f(K)$, there is an index $\alpha\in A$ such that $g(x)\in U_\alpha$. By the definition of inverse image, this is exactly the statement that $x\in g^{-1}(U_\alpha)$. Hence
\begin{align*}
K\subset \bigcup_{\alpha\in A}g^{-1}(U_\alpha).
\end{align*}
Now we use compactness of $K$. Since $(g^{-1}(U_\alpha))_{\alpha\in A}$ is an open cover of $K$, there exist finitely many indices $\alpha_1,\dots,\alpha_m\in A$ such that
\begin{align*}
K\subset \bigcup_{j=1}^{m}g^{-1}(U_{\alpha_j}).
\end{align*}
This finite family is the pulled-back finite subcover that we will push forward to cover $f(K)$.
[/guided]
[/step]
[step:Push the finite subcover forward to cover $f(K)$]
We claim that the finite family $U_{\alpha_1},\dots,U_{\alpha_m}$ covers $f(K)$. Let $y\in f(K)$. Since $g:K\to f(K)$ is surjective, there exists $x\in K$ such that $g(x)=y$. From
\begin{align*}
K\subset \bigcup_{j=1}^{m}g^{-1}(U_{\alpha_j}),
\end{align*}
there exists $j\in\{1,\dots,m\}$ such that $x\in g^{-1}(U_{\alpha_j})$. Therefore $y=g(x)\in U_{\alpha_j}$. Hence
\begin{align*}
f(K)\subset \bigcup_{j=1}^{m}U_{\alpha_j}.
\end{align*}
Thus every open cover of $f(K)$ has a finite subcover, so $f(K)$ is compact.
[/step]
[step:Apply the same argument to the inverse homeomorphism]
Assume that $f(K)$ is compact in the subspace topology inherited from $Y$. The inverse map $f^{-1}:Y\to X$ is a homeomorphism because $f$ is a homeomorphism. Applying the forward implication already proved to the homeomorphism $f^{-1}:Y\to X$ and to the subset $f(K)\subset Y$, we obtain that
\begin{align*}
f^{-1}(f(K))
\end{align*}
is compact in the subspace topology inherited from $X$. Since $f$ is injective and $K\subset X$, we have
\begin{align*}
f^{-1}(f(K))=K.
\end{align*}
Therefore $K$ is compact. Combining this with the forward implication proves that $K$ is compact if and only if $f(K)$ is compact.
[/step]
