Compactness Invariant between normed spaces Let $X$ and $Y$ be finite dimensional normed spaces. Let $D:\X \rightarrow Y$ be an isometric isomorphism then if $X$ is compact the $Y$ is also compact.

I have started by choosing a sequence in $Y$ and then taking its inverse image but I am stuck.

Any hints are greatly appreciated!
Chival mentioned in the comments that a normed space cannot be bounded (since it is a vector space).
Rephrasing the question to overcome this issue, we get


  Let $X$ and $Y$ be finite dimensional normed spaces. Let $D:X \to Y$ be an isometric isomorphism. Prove that if $Z \subset X$ is compact then $D(Z)$ is also compact.


Then, by noticing $D$ is continuous, the conclusion follows, because as stated by Frank in another answer the following theorem holds:


  If $X$ and $Y$ are topological spaces and $f: X \rightarrow Y$ is continuous, and if $X$ is compact, then $Y$ is compact.
Look up the following (more general) theorem: If $X$ and $Y$ are topological spaces and $f: X \rightarrow Y$ is continuous, and if $X$ is compact, then $Y$ is compact.