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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,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 andThe continuous product function makes $F$ closed, but $\pi (F)=\mathbb{R}\setminus \{0\}$ is not closed. Thus compactness of the discarded factor is essential to the general conclusion.

For metric spaces $X,Y$ with $X$ compact, the projection with a compact factor is closed: if $F\subset X\times Y$ is a closed set and $y_n\in {\pi }_Y(F)$ converges to $y$, choose $x_n$ with $(x_n,y_n)\in F$. Sequential compactness supplies $x_{n_k}\to x$, so $(x_{n_k},y_{n_k})\to (x,y)\in F$, proving $y\in {\pi }_Y(F)$. The projection map is also continuous for the sum metric. Without compactness, the closed set $\{(x,y)\in {\mathbb{R}}^2:xy=1\}$ projects to the nonclosed set $\mathbb{R}\setminus \{0\}$.