Am I right understanding a least point theorem in general vector space? The theorme is as follows

Let $(V,\|\cdot\|)$ be a normed space, and let $W$ be a finite-dimensional subspace of $V$. then there is at least one closest point $w^*\in W$ to $v$.
  that is there is $w^*\in W$ such that $\|v-w^*\|=\inf\{\|v-w\|:w \in W\}$

Proof is introduced as follows
Notice that the zero vector is in $W$, and so 
$$
\inf\{\|v-w\|:w \in W\} \le \|v-0\|=\|v\|
$$
Let $M=\|v\|$. If $w$ satisfies $\|v-w\| \le \|v\|$, then
$$
\|w\| \le \|w-v\|+\|v\| \le M+M=2M
$$
Thus if we define $K :=\{w \in W : \|w\| \le 2M \}$,then
(now, my curious part is here)
$$
**\inf\{\|v-w\| : w \in K \} = \inf\{\|v-w\| : w \in W\}**
$$
the book prove subsequently without any explanation. So I was trying to fill the skipped part. Please check my thought. 
Let
$$\rho_{1}(w)=\inf\{\|v-w\| : w\in W\} $$
$$ \rho_{2}(w)=\inf\{\|v-w\| : w \in K\}$$
then by above 
$$0 \le \rho_{1}(w), \rho_{2}(w)\le M$$
since we can select $M \ge 0$ by choosing arbitrary $v \in V$, then for all $M \ge 0$
$$
0\le |\rho_{1}(w)-\rho_{2}(w)| \le M 
$$ 
Thus 
$$\rho_1(w)=\rho_2(w)$$ the proof is finished. 
I want to check my solution is right and if there is another solution above, please give me a information.
 A: Since $K \subseteq W$, we have:
$$\inf\{\|v-w\| : w \in K \} \geq \inf\{\|v-w\| : w \in W\}$$
On the other hand, if $w \in W$ is such that $\|v-w\|\leq \|v\|$, then by the above argument we have $w \in K$. Thus, $\{w \in W : \|v-w\|\leq \|v\|\} \subseteq K$ so:
\begin{align}\inf\{\|v-w\| : w \in W\} &= \inf\{\|v-w\| : w \in W\text{ such that } \|v-w\|\leq \|v\|\}\\ &\geq \inf\{\|v-w\| : w \in K \}\end{align}
Thus: 
$$\inf\{\|v-w\| : w \in K \} =\inf\{\|v-w\| : w \in W\}$$
Regarding your proof, you probably mean to write
$$\rho_{1}(v)=\inf\{\|v-w\| : w\in W\}$$
$$\rho_{2}(v)=\inf\{\|v-w\| : w\in K\}$$
since $w$ is only a dummy variable inside the set; the real dependence is on $v$.
You actually showed that $\lim_{v\to 0}|\rho_1(v) 
- \rho_2(v)| = 0$, which is in fact easy to see since as $v\to 0$ you will eventually have $\|v\|\leq 2M$.
However, this is not what you had to prove: you should prove that for a fixed $v \in V$ we have $\rho_1(v) = \rho_2(v)$.
A: Your solution doesn't work, because the vector $v$ is given here, so you can can't choose it at will, so as to satisfy such and such conditions.
Here is short proof of the assertion:
As you observed, $\;\inf_\limits{w\in W}\|v-w\|\le M $. Now
$$\inf_\limits{w\in W}\|v-w\|=\inf_\limits{w\in K}\|v-w\|$$
simply results from the fact that, if $w\in W\smallsetminus K$, i.e. if $\|w\|>2M$, then $\;\|v-w\|>M$.
Indeed, by the second triangle inequality,
$$\|v-w\|\ge\bigl|\|w\|-\|v\|\bigr|=\bigl|\|w\|-M\bigr|>2M-M.$$
