 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 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 $ϵ>0$, both coordinate distances are eventually less than $ϵ/2$, so their sum is less than $ϵ$. Hence the same subsequence converges in the given product metric to $(x,y)$. This proves the product is sequentially compact.