A counter example of best approximation Construct a point $f\in C[0,1]$ and a closed subspace $V\subset C[0,1]$ such that $f$ does not have a best approximation in $V$.
Definition:
$C[0,1]$ is the set of countinous function with the norm $||f||=\max\limits_{x\in[a,b]}|f(x)|$.
Let $B$ be a normed vector space with norm $||\cdot||$. Set $V\subset B$ and $y_0\in B$.  A vector $x_0\in V$ is called the best approximation of $y_0$ in V if 
$$||y_0-x_0||=\operatorname{dist}(y_0,V)=\inf_{x\in V}||y_0-x||$$
 A: One can use the following general fact (whose proof is not difficult): if $\Phi$ is any continuous linear functional on $B$ and if $f\in B\setminus \ker(\Phi)$, then 
$$\Vert\Phi\Vert=\frac{\vert\Phi(f)\vert}{{\rm dist}(f,\ker(\Phi))}\, . $$
It follows that if you start with any linear functional $\Phi$ which does not attain its norm (i.e. there is no $x$ with $\Vert x\Vert=1$ and $\Phi(x)=\Vert\Phi\Vert$), then $V=\ker\Phi$ and any $f$ such that $\Phi(f)\neq 0$ do the job.
In the special case you are looking at, $V=\mathcal C([0,1])$, consider the linear functional $\Phi$ defined by $$\Phi(x)=\int_0^{1/2} x(t)\, dt-\int_{1/2}^1 x(t)\, dt\, . $$
It is not hard to check that $\Vert\Phi\Vert=1$ and that $\Phi$ does not attain its norm (the idea is that if $x\in\mathcal C([0,1])$ satisfied $\Vert x\Vert=1=\Phi(x)$, then $x$ would have to be equal to $1$ on $(0,1/2)$ and to $-1$ on $(1/2,1)$, so $x$ would not be in the space $\mathcal C([0,1])$.) 
In conclusion, you may take
$$V=\left\{ x\in\mathcal C([0,1]);\; \int_0^{1/2}x(t)\, dt=\int_{1/2}^1x(t)\, dt \right\}$$
and, for example, $f(t)=t$.
