Let $k\in\mathbb N$, $H_i$ be a (finite-dimensional, if necessary) $\mathbb R$-Hilbert space for $i\in I:=\{1,\ldots,k\}$, $H:=\bigotimes_{i\in I}H_i$ denote the tensor product$^1$ of $(H_i)_{i\in I}$ and $$H^{(r)}:=\left\{u\in H:\operatorname{rank}u=r\right\}\;\;\;\text{for }r\in\mathbb N_0.$$

I'm struggling to understand the importance and implication of the following result: Let $v\in H$.

1. There is a $u\in H$ with $\operatorname{rank}u=1$ and $$\left\|u-v\right\|_H=\inf_{u\in H^{(1)}}\left\|u-v\right\|_H\tag1.$$
2. There is a $u\in H^{(3)}$ and a $(u_n)_{n\in\mathbb N}\subseteq H^{(2)}$ with $$\left\|u_n-u\right\|_H\xrightarrow{n\to\infty}0\tag2.$$

Okay, by 1., there is a (not necessarily unique) minimizer of $$H^{(1)}\to[0,\infty)\;,\;\;\;u\mapsto\left\|u-v\right\|_H\tag3.$$

Question 1: But why can we infer from 2. that the analogous problem of minimizing $$H^{(2)}\to[0,\infty)\;,\;\;\;u\mapsto\left\|u-v\right\|_H\tag4$$ may have no solution? I guess we need to take $v=u$ (with $u$ as in 2.), but why does the existence of $(u_n)_{n\in\mathbb N}$ imply that there is no solution?

Question 2: That we can only guarantee the existence of a minimizer of $$H^{(r)}\to[0,\infty)\;,\;\;\;u\mapsto\left\|u-v\right\|_H\tag5$$ for $r=1$ is unsatisfactory only if it would actually be beneficial to take $r$ as large as possible. I could imagine that the error $\left\|u^{(r)}-v\right\|_H$ of a hypothetical minimizer $u^{(r)}$ of $(5)$ is nonincreasing in $r\in\mathbb N_0$. Is this the case? If so, how can we show this?

Question 3: Can we infer from 2. that $(5)$ may have no minimizer for all $r\ge2$?

$^1$ If $E_i$ is a $\mathbb R$-vector space, I'm defining $$(x_1\otimes x_2)(B):=B(x_1,x_2)\;\;\;\text{for }B\in\mathcal B(E_1\times E_2)\text{ and }x_i\in E_i,$$ where $\mathcal B(E_1\times E_2)$ is the space of bilinear forms on $E_1\times E_2$, and $$E_1\otimes E_2:=\operatorname{span}\{x_1\otimes x_2:E_i\in E_i\}\subseteq{\mathcal B(E_1\times E_2)}^\ast.$$