Given a point $x$ and a closed subspace $Y$ of a normed space, must the distance from $x$ to $Y$ be achieved by some $y\in Y$?

I think no. And I am looking for examples. I would like a sequence $y_n$ in $Y$ such that $||y_n-x||\rightarrow d(x,Y)$ while $y_n$ do not converge.

Can anyone give a proof or an counterexample to this question?

This is a slight adaptation of a fairly standard example.

Let $\phi: C[0,1]\to \mathbb{R}$ be given by $\phi(f)=\int_0^{\frac{1}{2}} f(t)dt - \int_{\frac{1}{2}}^1 f(t)dt$. Let $Y_\alpha = \phi^{-1}\{\alpha\}$. Since $\phi$ is continuous, $Y_\alpha$ is closed for any $\alpha$.

Now let $\hat{f}(t) = 4t$ and notice that $\phi(\hat{f}) = -1$ (in fact, any $\hat{f}$ such that $\phi(\hat{f}) = -1$ will do). Then $$\inf_{f \in Y_0} \|\hat{f}-f\| = \inf \{ \|g\|\, | \, g+\hat{f} \in Y_0 \} = \inf \{ \|g\|\, | \, \phi(g) =1 \} = \inf_{g \in Y_1} \|g\|$$ It is clear that $g_n$ is an infimizing sequence for the latter problem iff $g_n+\hat{f}$ is an infimizing sequence for the initial problem.

It is well known that $Y_1$ has no element of minimum norm, consequently there is no $f \in Y_0$ that mnimizes $\|f-\hat{f}\|$.