= Solution
{c}
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 \b[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 therefore
$$
d((x_{n_{k_j}},y_{n_{k_j}}),(x,y))
=d_X(x_{n_{k_j}},x)+d_Y(y_{n_{k_j}},y)\to0.
$$
Thus \b[$X\times Y$ 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.
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