Show that $||x-y|| > 1/2$ for all $y \in Y$. I'm sorry for posting this question, but I really have no clue on how to proceed:


*

*Let $E$ be a normed vector space.  If $Y$ is a closed proper subspace of $E$, then there is a $x \in E$ such that $||x|| = 1$ and $||x - y|| > 1/2$ for all $y \in Y$.


*If $E$ is of infinite dimension, then the unit ball $B_1 = \{ \, x \in E : ||x|| \le 1 \, \}$ is never compact in strong topology.

For statement 1, I attempted using Hahn-Banach to find a $g:  E \rightarrow \mathbb R$ such that:
$$g(x) > \alpha > g(y)$$
Since $Y$ is a subspace, it is also true that $g(x) > \alpha > g(cy) = cg(y)$. Thus, $g = 0$ on $Y$.  Then, because $g$ is a linear functional, we can re-scale it such that $||x|| = 1$ and the above relation still holds true.  This gives us:
$$g(x - y) > \alpha = \epsilon/2 \, \, \rightarrow \, \, g(x - y) > 1/2$$
Then, we have:
$$||x-y|| = \sup_{||f|| \le 1} |f(x-y)| \ge g(x - y) > 1/2$$
Well, the above requires that $||g|| = \sup \{ \, |g(x)| : ||x|| \le 1 \, \} \le 1$, something I cannot justify ....   Then for statement 2, I have absolutely no clue.
 A: For the second one, we can use a nice result called the Riesz lemma:

Let $X$ be a non-dense closed subspace of $Y$. Let $0 < r < 1$. There exists $y \in Y$ such that $||y||=1$ and $\inf_{x \in X} ||x-y|| > r$.

Now, let $x \in B$, where $B$ is the unit ball. Then there exists $x_2$ satisfying Riesz' Lemma for $r=0.5$ and the subspace generated by $x_1$. Next, there exists $x_3$ satisfying Riesz' Lemma for $r=0.5$ and the subspace generated by $x_1 ,x_2$. You can proceed to construct a sequence like this. 
Note that for all $x_i, x_j \in \{ x_n\}$, $|x_i-x_j| > r$. Hence no subsequence of $\{x_n\}$ is Cauchy, hence $\{x_n\}$ is a sequence with no convergent subsequence. This proves that $B$ is not sequentially compact, hence compact.
The key to the above proof is that every finite dimensional space is not dense in an infinite dimensional space. (In fact, it is nowhere dense).
For the first question, you can again use Riesz lemma. Since $Y$ is not dense in $E$ (it's closure is $Y$ which is not $E$), by the Riesz Lemma, there exists for $r=0.5$, a point $x$ such that $\inf_{y \in Y} ||x-y|| > 0.5$ and  $||x||=1$. Which directly finishes your question.
To prove the Riesz Lemma, note that if $X$ is a non-dense closed subspace of $Y$, then for any $y_1 \notin X$, note that $R=\inf_{x \in X} ||x-y_1|| > 0$. For $\epsilon>0$, let $x_1 \in X$ be such that $||x_1-y_1|| < R+\epsilon$. Let $y_0 = \frac{y_1-x_1}{||y_1-x_1||}$, then $||y_0||=1$, furthermore:
$$
\inf_{x \in X} ||x-y||  = \inf_{x \in X} \Bigg|\Bigg|x+\frac{x_1}{||x_1-y_1||} - \frac{y_1}{||x_1-y_1||}\Bigg|\Bigg| > \frac{R}{R + \epsilon}
$$
Hence, we can let $\epsilon$ be as small as we want, and get $\frac{R}{R + \epsilon}$ greater than any $1>r>0$ we want.
