= Solution
Assume the subsequence property, and let $f:E\to\mathbb R$ be a <continuous function>. If $f$ were unbounded, for each positive integer $j$ we could choose $x_j\in E$ with $|f(x_j)|>j$. There is a subsequence $x_{j_k}\to x$ with $x\in E$. By <continuity>, $f(x_{j_k})\to f(x)$, so that real sequence is bounded. But $|f(x_{j_k})|>j_k\to\infty$, a contradiction. Hence \b[every continuous real-valued function on $E$ is bounded]. This is the bounded-function consequence of <sequential compactness>.
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