About an extension of Riesz' Lemma for normed spaces The Riesz' Lemma is as follows: 
Let $Y$ and $Z$ be subspaces of a normed space $X$ of any dimension (finite or infinite) such that $Y$ is closed (in $X$) and is also a proper subset of $Z$. Then for every real number $\theta$ in the open interval $(0,1)$, there is a point $z$ in $Z$ such that $$||z|| = 1$$ and $$||z-y|| \geq \theta$$ for every $y$ in $Y$. 
Now we want to prove the following assertion: 
If $Y$ is finite-dimensional, then we can even take $\theta$ to be equal to $1$ in the statement of the Riesz' Lemma. 
How to prove this? 

I would apply the following trick in case $\dim Y<\infty$:
Let $z_0\in Z\setminus Y$ arbitrary and imagine the finte dimensional normed space $U$ spanned by $\langle Y,z_0\rangle$. Then there are many ways to continue, for example, the unit ball is compact in $U$, thus applying Riesz's lemma to $\theta_n:=1-\frac1n$ and $Y\subset U$, we get a sequence $u_n$ with $||u_n||=1$ and $d(u_n,Y)>1-\frac1n$. Pick a convergent subseqence and let $z$ be its limit.
