= Solution
{c}
Both properties are preserved by a <continuous map>.
For <sequential compactness>, take any <sequence> $(y_n)$ in $f(X)$ and choose $x_n\in X$ with $f(x_n)=y_n$. If $X$ is <sequentially compact>, some <subsequence> satisfies $x_{n_k}\to x\in X$. By <continuity>, $y_{n_k}=f(x_{n_k})\to f(x)\in f(X)$. Hence $f(X)$ is a <sequentially compact space>.
For <connectedness>, suppose instead that $f(X)=U\cup V$ is a separation into disjoint nonempty relatively <open sets>. Then $f^{-1}(U)$ and $f^{-1}(V)$ are disjoint nonempty <open sets> of $X$ whose union is $X$, contradicting <connectedness>. Therefore $f(X)$ is a <connected space>.
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