Properties of 1-Sphere in a linear normed space against a normed linear subspace. Hi i have a little problem in understanding the proof of the following theorem
If $N$ is a finite-dimensional proper subspace of normed
linear space $X$, there exists an element in the 1-sphere of $X$ whose distance
from $N$ is 1.
The proof basically takes a point $z$ on $X$ but not in $N$ so there exists a sequence $(n_k)$ of points in $N$ such that $||z - n_k|| \rightarrow d(z,N)$ (why?) ssince $N$ is finite dimensional and $n_k$ is bounded (why?) there exists a subsequence $v_k$ of $n_k$ which converges to some $n \in N$. Hence
$||z - n|| = \lim_{k \rightarrow +\infty} ||z - v_k|| = d(z,N) = d(z-n,N)$ (why $d(z,N) = d(z - n,N)$?).
The remaining part it is easy to understand...
 A: Since $z\not\in N$ and $N$ is closed, necessarily $\text{dist}(z,N)>0$ (if the distance is zero, it means that $z$ is in the closure of $N$). 
By definition,
$$
\text{dist}(z,N)=\inf\{\|z-n\|:\ n\in N\}.
$$
So the numbers $\|z-n\|$ can be arbitrarily close to $\text{dist}(z,N)$, i.e. we are allowed to choose a sequence $\{n_k\}_k\subset N$ with 
$$
\lim_{k\to\infty}\|z-n_k\|=\text{dist}(z,N).
$$
As the sequence of numbers $\{\|z-n_k\|\}_k$ is convergent, it is bounded; there exists $c>0$ with $\|z-n_k\|\leq c$ for all $k$. 
Then
$$
\|n_k\|\leq\|z-n_k\|+\|z\|\leq c+\|z\|,
$$
which shows that the sequence $\{n_k\}$ is bounded. Since $N$ is finite-dimensional, closed and bounded sets are compact. So the sequence $\{n_k\}_k$ admits a convergent subsequence. If $n=\lim n_k\in N$ (remember that $N$ is closed, since it is finite-dimensional),
$$
\|z-n\|=\lim_k\|z-n_k\|=\text{dist}(z,N).
$$
The first equality is simply the continuity of the norm: using the reverse triangle inequality,
$$
\left|\,\|z-n\|-\|z-n_k\|\,\right|\leq\|(z-n)-(z-n_k)\|=\|n-n_k\|\to0.
$$
Finally, I don't see why it is needed, but $\text{dist}(z,N)=\text{dist}(z-n,N)$ because $n\in N$ and $N$ is a subspace:
$$
\text{dist}(z,N)=\inf\{\|z-r\|:\ r\in N\}=\inf\{\|z-n-r\|:\ r\in N\}=\text{dist}(z-n,N);
$$
as $N$ is a subspace, the elements $n-r$, as $r$ goes through all of $N$, describe all of $N$. 
