Let $C$ be a centrally-symmetric closed convex subset of $\mathbb R^n$. For example, think of $C$ as the unit-ball for an $\ell_p$-norm, for some $p \in [1,\infty]$. Let $x=(x_1,\ldots,x_n) \in \mathbb R^n$ and $y = (y_1,\ldots,y_n) \in C$ be the orthogonal projection of $x$ onto $C$.

Question. Is it true that $|y_i| \le |x_i|$ for all $i$ ?

The above claim "appears true", at least when one draws a diagram (of course this is itself not a proof).

My attempted proof

Fix an index $i \in [n]$. If $y_i = x_i$, there is nothing to show. Otherwise, suppose $x_i \ge 0$. Then because $x_i$ is symmetric w.r.t the $i$ axis, we must have $y_i \ge 0$, since $C$ is symmeric about the $i$th axis of $\mathbb R^n$. Also, we must $x-te_i \in C$ for sufficiently small positive $t$ (otherwise $y_i=x_i$). Here $e_i$ is the $i$th standard basis vector in $\mathbb R^n$. Now, by the Kolmogorov characterization of projections, we have

$$ 0 \ge (x-y)^\top (x-te_i-x) = -tx_i+ty_i, $$ that is, $0 \le y_i \le x_i$ as claimed. Similarly, if $x_i \le 0$, then use $x+te_i$ instead of $x-te_i$ to get $x_i \le y_i \le 0$. The result would then follow.

However, I'm not 100% about my proof above. Thanks in advance.