Distance between elements with norm 1 and subset of a closed subset. Let $I=[0,1]$ be the unitary interval, let $X$ be a closed subset of $C(I)$ whose elements are the continuous function vanishing at zero, let $Y$ be the subset of $X$ such that $\int_{0}^{1}x(t)dt=0$, prove that for all $x\in X$ such that $||x||=1$ hold that $dist(Y,x)<1.$ 
I define the fuction $u(x)=\int_{0}^{1}x(t)dt$ the kernel of this map is $Y$ and the projection map is such that $||p(x)||_{X/Y}=dist(Y,x),$ BUT i don´t see how finish the proof, well I know that there are $u_o:X/Y \rightarrow \mathbb{R}$ which is an isomorphism and the norm of $u$ and $u_o$ are the same, some body can help me to finish... 
 A: I think proving this directly is also possible. Let $x\in X$ be given with $||x||=1$. I will construct $y\in Y$ explicitly. Define 
$$A=\{t\in I:x(t)\geq 0.5\}$$
$$C=\{t\in I:x(t)\leq-0.5\}$$
$$B=I\setminus A\setminus C$$
Now $B$ is a finite union of intervals. Define the first interval to be $[0,\delta)$ (it must start from $0$ by $x\in X$). 
Define $y(t)=-\epsilon$ for $t\in C$. 
Define $y(t)=\epsilon$ for $t\in A$. 
Define $y(t)$ on $B\setminus[0,\delta)$ as linearly from $-\epsilon$ to $\epsilon$, to make $y(t)$ is continuous. Note that these $t$'s do not contribute to the integral since these pieces are symmetric. 
We now have to make up for the $\epsilon(\left|A\right|-\left|C\right|)$ in $[0,\delta)$. To be able to do so, let $\epsilon$ be small enough that $\frac{\epsilon\left|\left|A\right|-\left|C\right|\right|}{\delta}<\frac{1}{4}$. Hopefully you will be able to fill in the details of defining $y(t)$ on $[0,\delta)$ such that it has integral $0$, being only allowed to take values inside $[-1/2,1/2]$, having $y(0)=0$ and $y(\delta)=\pm\epsilon$. 
