= Solution
The sum of the two <metrics> is nonnegative and symmetric. It vanishes exactly when both coordinate distances vanish, hence exactly when the two points agree. Adding the two component <triangle inequalities> gives the <triangle inequality> on the <Cartesian product>, so the displayed function is a <metric>. Moreover,
$$
d_Y(\pi(x_1,y_1),\pi(x_2,y_2))\le d((x_1,y_1),(x_2,y_2)),
$$
so the <projection map> is $1$-<Lipschitz> and therefore <continuous>.
To prove <sequential compactness> directly from <compactness>, take a sequence $(x_n)$ in $X$. If every point $x$ had a neighborhood containing sequence terms at only finitely many indices, these neighborhoods would form an open cover with a finite subcover, implying that only finitely many sequence terms exist, a contradiction. Thus some $x\in X$ has infinitely many indices in every neighborhood. Choose increasing indices $n_k$ with $d_X(x_{n_k},x)<1/k$; then $x_{n_k}\to x\in X$.
Now let $F\subset X\times Y$ be <closed> and suppose $y_n\in\pi(F)$ converges to $y$. Choose $x_n$ such that $(x_n,y_n)\in F$. A <convergent subsequence> $x_{n_k}\to x$ exists by the preceding argument. In the product <metric>, $(x_{n_k},y_{n_k})\to(x,y)$, and the fact that $F$ is <closed> gives $(x,y)\in F$. Thus $y\in\pi(F)$, proving that the <projection with a compact factor is closed>. The sequential criterion for a <closed set> in a <metric space> follows by choosing points at distance less than $1/k$ from any point of its closure.
For a counterexample without a compact factor, take $X=Y=\mathbb R$ with their usual <metrics> and
$$
F=\{(x,y):xy=1\}.
$$
The continuous product function makes $F$ <closed>, but $\pi(F)=\mathbb R\setminus\{0\}$ is not <closed>. Thus \b[compactness of the discarded factor is essential to the general conclusion].
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