 Solution

A metric space is sequentially compact if every sequence in it has a subsequence converging to a point of the same space. The sequence $x_n=1/(n+1)$ lies in $(0,1)$, but every subsequence converges in $\mathbb{R}$ to zero. Uniqueness of limits excludes an alternative limit inside $(0,1)$, so the open interval is not sequentially compact.

For the product, take any sequence $(x_n,y_n)$. sequential compactness of $X$ gives a subsequence $x_{n_k}\to x\in X$. Apply sequential compactness of $Y$ to the corresponding sequence $y_{n_k}$ to obtain a further subsequence $y_{n_{k_j}}\to y\in Y$. Passing to a subsequence preserves the convergence of the first coordinate, and thereforeThus $X\times Y$ is sequentially compact with the stated sum metric. Taking the second coordinate subsequence inside the first one is essential; two unrelated subsequences would not suffice.