If $X$ is reflexive and $A \in \mathcal{L}(c_0,X)$ then show that $A$ is a compact operator. Exercise :

Let $X$ be a reflexive Banach space and $A \in \mathcal{L}(c_0,X)$. Show that $A$ is a compact operator.

Thoughts :
First of all, I know that if $X,Y$ are Banach then $A \in \mathcal{L}_c(X,Y)$ iff $A^* \in \mathcal{L}_c(X,Y)$. In our case, $c_0$ is a banach space (when equipped with the sup-norm). So essentialy it may come down to proving such a statement. 
If the adjoint operator $A^* : X^* \to c^*_0$ is considered, then I know that $c_0^* \simeq \ell_1$ and that $X^*$ should be reflexive too, but I seem to struggle using sequential statements to work a proof.
Also, this is part of exercises on our notes after being introduced to Spectral Theory, so there may be a solution involving that, despite not having found anything that may help (as most Lemmas involve a Bounded Linear Functional from a space to itself).
Any hints or elaborations will be highly appreciated.
 A: 
An operator $T$ from a Banach space $X$ to a Banach space $Y$ is compact if      and only if for every bounded sequence $x_n$ in $X$, there is a subsequence $x_{n_k}$ such that $Tx_{n_k}$ converges in $Y$ (in the norm).

Consider the adjoint operator $A^*:X^*\to (c_0)^*\simeq\ell_1$. Let $(x_n^*)$ be a bounded sequence in $X^*$. Since $X$ is reflexive, so is $X^*$. Every bounded sequence in a reflexive Banach space has a weakly convergent subsequence, so there exists a subsequence $(x_{n_k}^*)$ converging weakly.
Since $A^*$ is continuous, it is also weakly continuous, that is, $A^*x_{n_k}^*$ converges weakly to some element $y\in\ell_1$. But in $\ell_1$ weak convergence is equivalent to norm-convergence. Therefore $A^*x_{n_k}^*$ converges in norm. We have thus verified that the condition above holds, so $A^*$ is compact, and so is $A$.
Added in response to some requests for clarification.
The argument above relies on some classical theorems in Banach space theory, which apparently are not sufficiently well known.
The first theorem is that in a reflexive Banach space every bounded sequence has a weakly convergent subsequence. For a proof, see here.
The second theorem is that in $\ell_1$ the weak and norm-topologies are equivalent. In particular, weak convergence of sequences is equivalent to norm convergence of sequences. Most textbooks on Banach spaces contain this theorem. For several references, see here.
