a subset of $l_2$.

Consider $xin A$. Since for all $ige 1$ we have $x_i^2le (i+1/i)x_i^2$ we conclude by adding these inequalities that $Vert xVert_2^2le 1$, so $$sup_{xin A}Vert xVert_2le 1.$$ Now, if we define $x^{(k)}inell_2$ by $x^{(k)}_k=1/sqrt{1+1/k}$ and $x^{(k)}_i=0$ for $ine k$, then $x^{(k)}in A$ and $$Vert x^{(k)}Vert_2=frac{1}{sqrt{1+1/k}}le sup_{xin A}Vert xVert_2le 1$$ But since $k$ is arbitrary we conclude by letting $k$ tend to $infty$ that $$ sup_{xin A}Vert xVert_2=1.$$

Now, suppose that there exists $yin A$ such that $Vert yVert_2=1$ then $$sum_{ige1}frac1i y_i^2= sum_{ige1}left(1+frac1i right)y_i^2 -sum_{ige1}y_i^2le1-Vert yVert_2^2=0$$ Thus $$sum_{ige1}frac1i y_i^2=0$$ So $y_i=0$ for all $i$ and this contradicts the fact that $Vert yVert_2=1$. This contradiction proves that there is no $yin A$ such that $Vert yVert_2=1$. $qquadsquare$