= Solution
Suppose for contradiction that the <continuous image of a connected space> $Y=f(X)$ is disconnected. Then there are disjoint nonempty sets $U,V$, open in the <subspace topology> on $Y$, with $Y=U\cup V$. By <continuity>, $f^{-1}(U)$ and $f^{-1}(V)$ are disjoint open subsets of $X$; by <surjective function>[surjectivity] both are nonempty, and their union is $X$. This is a <separation of a topological space> of $X$, contradicting that $X$ is a <connected space>. Hence $Y$ is connected.
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