It is true that every normed linear space $X$ is homeomorphic to its open unit ball $B$, for example via the homeomorphism $f:B\to X$ defined by $f(x)=\tan(\frac{\pi}{2}\|x\|)\cdot x$. This means that, as topological spaces, $X$ and $B$ are essentially identical. However, a normed linear space carries more information than a topological space, namely a norm. Thus the appropriate notion of "essentially identical" for normed linear space is the existence of an isometric isomorphism, a homeomorphism which preserves the norm. Clearly $X$ is not isometrically isomorphic to $B$, since $B$ is not even a normed linear space unless $X=\{0\}$. To see this, take any $x\neq 0$. Then $x/2\|x\|\in B$ but $2\cdot x/2\|x\|\notin B$.