I have tried this problem as: By using the definition
For a subspace $S$ of a normed linear space $X$, for all $x$ belongs to $X,g\in S$ is a best approximation of $x$ if $\|x-g\|=\inf\{\|x-g'\|:g'\in S\}$.
So in this case we have for $(x,y)\in\mathbb{R}^2 , \|(x,y)-(x',x')\|=\inf\{\|(x,y)-(y',y')\|\colon(y',y')\in B\}$. I don't know what to do after this ? Should it be like $(x',x')=\mathbb{R}^2$ or anything else..kindly give me some idea.
Thank you in advance.
$(0,1)$ does not have infinitely many "best approximations" in the given subspace. For any given vector space $V$ and a subspace $H$, the vector $x \in V$, the vector in $H$ closest to $x$ is the projection of $x$ onto $H$, which is unique. For your case, you can see this geometrically. $B$ is the straight line $y = x$ in the plane. Now, there is only one point on said line closest to $(0,1)$, which is the point $\frac{\sqrt{2}}{2}(1,1)$.
You can also find this using the standard formula for an orthogonal projection from linear algebra. Observe that $B = \text{Span} \{(1,1)\}$. The projection is given by
$$\frac{(1,0)\cdot(1,1)}{|(1,1)|}(1,1) = \frac{\sqrt{2}}{2}(1,1)$$