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An <open cover> of a <topological space> is a collection of open subsets whose union is the whole space. A <compact space> is one for which every <open cover> admits a finite <subcover>. The <subspace topology> on $A\subseteq Y$ consists of sets $A\cap O$ with $O$ open in $Y$.
Let $\{U_j\}$ cover $f(X)$ by relatively open sets, and write $U_j=f(X)\cap O_j$ with $O_j$ open in $Y$. By <continuity>, $\{f^{-1}(O_j)\}$ is an open cover of $X$. A finite subcollection covers $X$ by <compactness>, and its corresponding $U_j$ cover $f(X)$. Thus \b[a continuous image of a compact space is compact].
For an image that is not closed, take $X$ to be a singleton and $Y=\{0,1\}$ with the <indiscrete topology>. Send the sole point to $0$. This map is continuous and $X$ is compact, while $f(X)=\{0\}$ is not closed in $Y$, whose only closed sets are $\varnothing$ and $Y$. A <Hausdorff> hypothesis on the target would exclude this example.
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