Let $X$ be a normed linear space and $Y$ be a finite dimensional subspace of $X$. Show that $\exists x \in X$ with $||x||=1$ and $d(x, Y)=1$.
My approach: Using Riesz lemma, for any $t \in (0, 1)\ \exists x_n$ with $||x_t||=1$ such that $d(x_t, Y) >t$ so if we take a sequence of $t$ going to $1,$ we have a corresponding sequence of $x_t$ each of unit norm, at corresponding distance more than $t$. If the sequence has a convergent subsequence, then since norm is a continuous map, the limit will also have unit norm and be at distance $1$ from $Y.$ But I am unable to show why a convergent subsequence must exist. Can someone help?
Edit: X is infinite dimensional
Best Answer
This is false. If $X$ is finite dimensional and $X=Y$ then $d(x,Y)=0$ for all $x$.
Answer for the edited version: Take any $x \notin Y$. Then $d(cx,Y)=|c|d(x,Y)$ so we can find $c$ such that $d(z,Y)=1$ where $z=cx$. Now there exist $(y_n) \subset Y$ such that $\|z-y_n\| \to d(z,Y)=1$. Since $\|y_n\| \leq \|z-y_n\|+\|z\|$ it follows that $(y_n)$ is a bounded sequence in $Y$. Hence it has a convergent subsequence $(y_{n_k})$. If the limit of this sequence is $y$ we get $\|z-y\|=1$. Now take $x=z-y$ to finish the proof.
I have used the comment by Daniel Fisher above to construct the is proof and I am thankful to him for the comment.