29595959

证明：(1)因为$X_0$为$X$的真子空间，于是存在${x_1} \in X\backslash {X_0}$，记$$d = \mathop {\inf }\limits_{x \in {X_0}} \left\| {x - {x_1}} \right\|$$

(2)因为$X_0$是闭的，故$d>0$，否则存在${x_n} \in {X_0}$，且$\left\| {{x_n} - {x_1}} \right\| \to 0$，再由$X_0$是闭的推出${x_1} \in {X_0}$矛盾

(3)不妨设$\varepsilon  < 1$，则有$\frac{d}{{1 - \varepsilon }} > d$，由下确界的定义知，存在${x_2} \in {X_0}$，使得\[\left\| {{x_2} - {x_1}} \right\| < \frac{d}{{1 - \varepsilon }}\]

(4)令${x_0} = \frac{{{x_1} - {x_2}}}{{\left\| {{x_1} - {x_2}} \right\|}}$，则$\left\| {{x_0}} \right\| = 1$，对于任何$x \in {X_0}$，注意到${x_2} \in {X_0}$，我们有

\begin{align*}
\left\| {x - {x_0}} \right\|& = \left\| {x - \frac{{{x_1} - {x_2}}}{{\left\| {{x_1} - {x_2}} \right\|}}} \right\| = \frac{1}{{\left\| {{x_1} - {x_2}} \right\|}}\left\| {\left( {\left\| {{x_1} - {x_2}} \right\|x + {x_2}} \right) - {x_1}} \right\|\\&
\ge \frac{1}{{\left\| {{x_1} - {x_2}} \right\|}} \cdot d > 1 - \varepsilon
\end{align*}