TITLE: Converse of Schauder's Theorem about compactness of adjoint operator
QUESTION [5 upvotes]: It is well known (also known as Schauder's Theorem) that if $X$ and $Y$ are normed spaces and $T:X\to Y$ is a linear and compact operator, then also $T^*:Y^*\to X^*$ is compact. The converse  is true if $Y$ is complete. 
So the natural question is: 

Is there an "easy" example that shows that we cannot drop the completeness of $Y$ for the converse implication?

In order to prove the converse, one usually applies the first implication to the bidual. Thus, a counterexample should be rooted in the subtle difference between "relatively compact" and "totally bounded", but I cannot wrap my head around it.
Any ideas are highly appreciated. Thank you in advance!

REPLY [2 votes]: Edit: This does not work.
Here is a proof using completeness of $Y$:
We have that $T^{**} : X^{**} \to Y^{**}$ is compact.
Thus, $i_Y \, T = T^{**} \, i_X$ is compact (where $i_Y : Y \to Y^{**}$ and $i_X : Y \to X^{**}$ are the canonical embeddings).
Hence, if $(x_n)$ is a bounded sequence,
$(T^{**} \, i_X \, x_n)$ has a convergent subsequence indexed by $(n_k)$.
Since $i_Y$ is an isometry, $T \, x_{n_k}$ is Cauchy in $Y$.