Isometry maps closed unit ball to closed unit balI

Question

Let $X$ and $Y$ be Banach spaces.

We say that a linear operator $T:X\to Y$ an isometry if for every $x\in X,$ we have $$\|Tx\| = \|x\|.$$

Question: Let the closed unit ball of $X$ be $$B_X=\{x\in X: \|x\|\leq 1 \}.$$ Is it true that an isometry maps closed unit ball onto closed unit ball, that is, $$T(B_X) = B_Y?$$

I think it is true, as isometry preserves distance. So I proceed to prove the equality.

Clearly every isometry is an injection. Therefore, $$T(B_X)\subseteq B_Y.$$

However, I have trouble proving $$B_Y\subseteq T(B_X).$$ Any hint would be appreciated.

Answer 1

No, consider an isometric embedding of finite dimensional spaces with different dimension.