I try solve exercise from Diestel:
If Y is a proper closed linear subspace of $l_p$ (1 $<$ p < $\infty$), then there is an x $\in$ $S_x$ so that distance $d(x, Y)$ $>=1$. I try solve this this problem using Riesz lemma(but it is helpful only for $d(x, Y)$ $=s<1$.
Thank you on all you help.
There are several ways to prove the existence of such an $x$. All ways that I can think of make essential use of the fact that the closed unit ball of a reflexive space is weakly compact. That should not be surprising, since in non-reflexive spaces, such an $x$ need not exist. I will give two arguments, both use (the second argument only marginally) the
Lemma: Let $E$ be a normed space, and $F\subset E$ a linear subspace. Then $p_F \colon E \to [0,+\infty)$ given by
$$p_F(e) = d(e,F) = \inf \{ \lVert e-f\rVert : f\in F\}$$
is a seminorm with kernel $\ker p_F = \overline{F}$ and $p_F(e) \leqslant \lVert e\rVert$ for all $e\in E$.
Proof: It is clear that $0 \leqslant p_F(e) \leqslant \lVert e\rVert$ since $\lVert\cdot\rVert$ attains only non-negative values, and $0 \in F$. Furthermore, we have
$$p_F(e) = 0 \iff \bigl(\forall \varepsilon > 0\bigr)\bigl(\exists f\in F\bigr)\bigl(\lVert e-f\rVert < \varepsilon\bigr) \iff \bigl(\forall \varepsilon > 0\bigr)\bigl(B_\varepsilon(e)\cap F\neq\varnothing\bigr)\iff e\in \overline{F}.$$
Hence $p_F(0\cdot e) = 0 = 0\cdot p_F(e)$, and for $\alpha \in \mathbb{K}\setminus \{0\}$ - where $\mathbb{K}$ is the scalar field - we have $f\in F \iff \alpha\cdot f \in F$ for all $f\in E$, so
\begin{align} p_F(\alpha\cdot e) &= \inf \{ \lVert \alpha\cdot e - f\rVert : f\in F\} \\ &= \inf \{ \lVert \alpha\cdot e - \alpha\cdot f\rVert : \alpha\cdot f\in F\} \\ &= \inf \{ \lVert \alpha\cdot e - \alpha\cdot f\rVert : f\in F\} \\ &= \inf \{ \lvert\alpha\rvert\cdot \lVert e-f\rVert : f\in F\} \\ &= \lvert\alpha\rvert\cdot \inf \{ \lVert e-f\rVert : f\in F\} \\ &= \lvert\alpha\rvert\cdot p_F(e). \end{align}
To see that the triangle inequality holds for $p_F$, fix $\varepsilon > 0$, and for $e_1,\,e_2 \in E$ pick $f_1,\,f_2\in F$ with
$$\lVert e_k - f_k\rVert \leqslant p_F(e_k) + \frac{\varepsilon}{2}$$
for $k \in \{ 1,2\}$. Then
$$p_F(e_1 + e_2) \leqslant \lVert (e_1 + e_2) - (f_1 + f_2)\rVert \leqslant \lVert e_1 - f_1 \rVert + \lVert e_2 - f_2\rVert \leqslant p_F(e_1) + p_F(e_2) + \varepsilon.$$
Since $\varepsilon > 0$ was arbitrary, the desired $p_F(e_1+e_2) \leqslant p_F(e_1) + p_F(e_2)$ follows.
Now consider the case where $F$ is a closed subspace of $E$ and let $\pi \colon E \to E/F$ be the canonical map. Since $F \subset \ker p_F$,
$$\lVert \pi(e)\rVert_{E/F} := p_F(e)$$
is well-defined, and by the lemma above, $\lVert\cdot\rVert_{E/F}$ is a norm. Endowing $E/F$ with that norm, we see that $\pi$ is continuous with $\lVert \pi\rVert \leqslant 1$.
The next important fact is that $\pi$ is an open map with
$$\pi(U_E) = U_{E/F},$$
where $U_X$ denotes the open unit ball of the normed space $X$, $U_X := \{ x\in X : \lVert x\rVert < 1\}$. The inclusion $\pi(U_E) \subset U_{E/F}$ follows directly from $\lVert\pi\rVert \leqslant 1$, and for the other inclusion note that $p_F(e) < 1$ means there is an $f\in F$ such that $\lVert e-f\rVert < 1$, i.e. $e-f \in U_E$, and $\pi(e-f) = \pi(e)$.
Finally, we come to the proof of existence. Suppose $E$ is a reflexive Banach space, and $F \subset E$ a closed subspace. Since the closed unit ball $B_E = \{e\in E : \lVert e\rVert \leqslant 1\}$ is weakly compact, and $\pi$ is continuous when we endow both spaces, $E$ and $E/F$, with the weak topology, it follows that $\pi(B_E)$ is a weakly compact, hence weakly closed, subset of $E/F$, and
$$U_{E/F} = \pi(U_E) \subset \pi(B_E) = \pi\bigl(\operatorname{cl}_w(U_E)\bigr) \subset \operatorname{cl}_w\bigl(\pi(U_E)\bigr) = \operatorname{cl}_w(U_{E/F}) = B_{E/F}.$$
Since $B_{E/F}$ is the smallest weakly closed set containing $U_{E/F}$, it follows that in fact
$$\pi(B_E) = B_{E/F},$$
that is, for every $e\in E$ with $p_F(e) = 1$, there is an $\tilde{e}\in E$ with $\pi(\tilde{e}) = \pi(e)$ and
$$1 = d(\tilde{e},F) = \lVert\pi(\tilde{e})\rVert_{E/F} \leqslant \lVert\tilde{e}\rVert \leqslant 1.$$
If $F$ is a proper closed subspace of $E$, then $E/F \neq \{0\}$, and there are $e\in E$ with $p_F(e) = 1$.
For the second argument, we note that for a proper closed subspace $F$ of a normed space $E$, there is an $e\in E$ with $d(e,F) = 1$ - that is the first marginal use of the lemma above. For $r > 0$, let $K_r(e) := \{ x\in E : \lVert x-e\rVert \leqslant r\}$ denote the closed ball with radius $r$ and centre $e$. If $E$ is a reflexive Banach space, then $K_r(e)$ is weakly compact for every $r \in (0,+\infty)$. For every $r > 1$,
$$M_r := K_r(e) \cap F$$
is non-empty, since $d(e,F) = 1$, and it is weakly compact, since $K_r(e)$ is weakly compact and $F$ is weakly closed. Also, for $1 < r < s$ we have $M_r \subset M_s$, hence the family $\{ M_r : r > 1\}$ has the finite intersection property, and therefore
$$M := \bigcap_{r > 1} M_r \neq \varnothing.$$
Now, by construction $M = \{ f\in F : \lVert e-f\rVert \leqslant 1\} = \{ f\in F : \lVert e-f\rvert = 1\}$ (the second because $d(e,F) = 1$), and hence for every $f\in M$, $x = e-f$ is an element of $E$ with $\lVert x\rVert = 1$ and - another marginal use of the lemma - $d(x,F) = d(e,F) = 1$.