I'm trying to prove that the line segment is the minimizer of the distance
$$d(x,y)=\inf l(\gamma),$$ where $x,y\in X$, $X$ is a Banach space, $\gamma$ is a path from $x$ to $y$ and $$l(\gamma)=\int_{0}^{1}\frac{|\gamma '(t)|}{1+|\gamma(t)|}dt.$$
You can suppose that $x=0$. In his book, Ekeland said that follows by radial symmetry. I'm not understand how use this fact, or how to prove by another way. Someone know how to solve it? Thanks for your attention.
The statement as you wrote it is false. For general $(x,y)$ the minimiser does not have to be a line segment. If $x = 0$ however we can show that the line segment must be a (not the) minimiser. (In the book the author has it written correctly.)
A counterexample: let $X = \mathbb{R}^2$ and $\|\cdot\|$ the $\ell_\infty$ norm. Let $x = (1,0)$ and $y = (1,1)$. Let $\gamma_1$ be the line segment connecting $x$ and $y$: $\gamma_1(t) = (1,t)$. We have $\dot{\gamma}_1(t) = (0,1)$ and so $\|\dot{\gamma}_1\| = 1$ and $\|\gamma_1\| = 1$ (since $t \leq 1$). So $l(\gamma_1) = \frac12$.
Now let $\gamma_2$ be piecewise defined. For $t\in (0,\frac12)$ let $\gamma_2(t) = (1 + t,t)$. For $t\in (\frac12,1)$ let $\gamma_2(t) = (2-t,t)$. We have that $\|\dot{\gamma}_2\| = 1$ again in the $\ell_\infty$ norm, while $\|\gamma_2(t)\| > \|\gamma_1(t)\| = 1$ for all $t \neq 0,1$. This implies that $l(\gamma_2) > l(\gamma_1)$. And so we see an example of a Banach space in which the line segment is not necessarily the minimiser between two points $x,y$.
Another counterexample: let $X = \mathbb{R}^2$ yet again; still equipped with the $\ell_\infty$ norm. Let $x = 0$ and $y = (1,0)$. As computed in the book the line segment $\gamma_1$ connecting $x,y$ has $l(\gamma_1) = \log 2$. Let $\gamma_2$ be the piecewise defined segment as follows. For $t\in (0,\frac12)$ let $\gamma_2(t) = (t,t)$. And for $t\in (\frac12,1)$ let $\gamma_2(t) = (t,1-t)$. We have that $\|\gamma_2(t)\| = t = \|\gamma_1(t)\|$ and that $\|\dot{\gamma}_2(t)\| = 1 = \|\dot{\gamma}_1(t)\|$ and hence $l(\gamma_1) = l(\gamma_2)$. This shows that the minimiser need not be unique (provided that $\gamma_1$ minimises the distance).
The radial symmetry alluded to, I suppose, is the following consequence of the convexity of the the Banach norm. (Though in my opinion "radial symmetry" is an awfully poor name for this property; the author perhaps is inspired by the case of a Hilbert space where the radial symmetry manifests more clearly according to our "Euclidean" intuitions.)
Claim $\|\dot{\gamma}(t)\| \geq \frac{\mathrm{d}}{\mathrm{d}t} \|\gamma(t)\| $ for a differentiable curve $\gamma:I\to X$ where $X$ is a Banach space.
Proof: By definition $$ \lim_{h\to 0} \frac1h \left(\|\gamma(t+h)\| - \|\gamma(t)\|\right) = \lim_{h\to 0} \frac1h \left( \|\gamma(t) + h \dot{\gamma}(t) + o(h)\| - \|\gamma(t)\|\right)$$ by triangle inequality and the scaling property of the Banach norm $$ \leq \lim_{h\to 0} \frac1h \left( h \|\dot{\gamma}(t)\| + o(h)\right) = \|\dot{\gamma}(t)\|$$ CQFD.
Hence $$ l(\gamma) = \int_0^1 \frac{\|\dot{\gamma}(t)\|}{1 + \|\gamma(t)\|} \mathrm{d}t \geq \int_0^1 \frac{1}{1 + \|\gamma(t)\|} \mathrm{d} \|\gamma(t)\| = \log(1 + \|\gamma(1)\|) - \log(1 + \|\gamma(0)\|) \\ = \log(1 + \|y\|) - \log(1+\|x\|)$$which gives an absolute lower bound to $l(\gamma)$. (By considering the reverse curve that runs from $y$ to $x$ we can actually improve the bound above slightly to $l(\gamma) \geq \left| \log\frac{1+\|y\|}{1+\|x\|}\right|$.) Since the line segment from $0$ to $y$ realises this lower bound, it must be a minimiser.
Incidentally, the above also shows that any solution to the equation $$ \|\dot{\gamma}\| = \frac{\mathrm{d}}{\mathrm{d}t} \|\gamma\| $$ will necessarily be a minimiser of the distance. (This, while sufficient, is not necessary: consider the case where $\|x\| = \|y\|$. By the mean value theorem a differentiable curve connecting $x$ and $y$ must have a point on which the right hand side vanishes. )