Is the projection of a closed set always a Borel set?

Question

I'm dealing with this article https://projecteuclid.org/euclid.acta/1485890981, which is about optimal transport viewed from a geometric point of view. To me it seems like it's given for good that the canonic projection $\pi:\mathbb{R}^d\times\mathbb{R}^d\to\mathbb{R}^d,\pi:(\mathbf{x},\mathbf{y})\mapsto\mathbf{x}$ of a closed set is σ-compact, in particular a Borel set. Thinking about the famous counterexample of a projection of a closed not being closed, it seems reasonable. Is it true?

Answer 1

This question was modified after I posted this answer. It has changed from a deep question to a trivial one. In the earlier version no space was specified and I thought the OP was considering a closed subset of a product of two arbitrary toplogical spaces. Please see the revised answer at the end. Any analytic set in $\mathbb R$ is the projection of a closed set in $\mathbb R \times X$ for some Polish space $X$. ( We can take $X$ to be $\mathbb N ^{\mathbb N}$ for example). Ref. Theorem 4.1.1 of "A Course On Borel Sets" by S. M. Srivastava. There exist analytic sets in $\mathbb R$ which are not Borel. Hence there exist closed sets whose projections are not Borel. If you insist on getting a closed set in $\mathbb R^{2}$ I don't have an immediate reference but I believe such sets exist in this case also.

Answer to the modified question: Let $C$ be closed in $\mathbb R \times \mathbb R$. We can write $C$ as $\cup_{n=1}^{\infty }K_n$ with each $K_n$ compact. If $p$ denotes the projection under consideration then $p(C)=\cup_{n=1}^{\infty }p(K_n)$ which is a countable union of compact sets, hence Borel.