= Solution
The sequence $x_n=1/(n+1)$ lies in the open interval and converges in the ambient real line to zero. Every <subsequence> also converges to zero, and uniqueness of limits prevents it converging to any point of the interval. Thus \b[the interval is not <sequentially compact>].
For a sequence $(x_n,y_n)$ in the product, <sequential compactness> of $X$ supplies indices $n_j$ with $x_{n_j}\to x\in X$. Apply <sequential compactness> of $Y$ to the corresponding $y_{n_j}$, extracting a further <subsequence> $y_{n_{j_\ell}}\to y\in Y$. The first coordinate still tends to $x$. For any $\epsilon>0$, both coordinate distances are eventually less than $\epsilon/2$, so their sum is less than $\epsilon$. Hence the same <subsequence> converges in the given product metric to $(x,y)$. This proves \b[the product is <sequentially compact>].
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