 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 every continuous real-valued function on $E$ is bounded. This is the bounded-function consequence of sequential compactness.