Let $V = \mathbb{R}^n$ with the inner product $\langle\cdot,\cdot\rangle$ and $U \subset V$ a vector subspace. Then for $v\in V$ and $x \in U$ the inequality$$ ||v-x|| \leq ||v-u||$$ is satisfied for all $u \in U$ if and only if $x = \text{P}_{V\to U}(v)$ ($x$ is the orthogonal projection from $v$ onto $U$).
Does this equivalence also hold for norms that are not induced by the/a inner product, e.g. the Chebyshev distance $||x||_\infty = ||(x_1,...,x_n)||_\infty:= \max\{|x_1|,|x_2|,...,|x_n|\}$? So here it would be $||v-x||_\infty \leq ||v-u||_\infty$ although we are still in an inner product space.
Let's assume $V$ would only be a normed vector space, $w$ might only be a best approximation. So the inequality still holds but the vector is not unique. But without an inner product does the term orthogonal projection still make sense? I think it doesn't, which is why multiple vectors would fit for $x$.