Proof check: the image of the closed unit ball under a compact operator is closed if the domain is reflexive. The task at hand is the following:
Let $K: X \to Y$ be a compact operator, $X$ and $Y$ Banach spaces, where $X$ is reflexive in addition. 
Claim: $K(B)$ is a closed subset of $Y$ where $B$ denotes the closed unit ball in $X$.
I'd like someone to go over the proof and point out mistakes, what could have been done better, faster or where I should have been more precise.
Proof:
Let $K(x_n)$ be a sequence in $K(B)$ (i.e. $x_n$ a sequence in $B$) that converges in $Y$. Then by the Eberlein theorem we obtain that every bounded sequence $x_n$ in $X$ has a weak convergent subsequence $x_{n_i}$ with weak limit $x \in X$. Since K is a compact operator we can then extract a subsequence $x_{n_{i_j}}$ such that $K(x_{n_{i_j}})$ converges to some $y \in Y$.
Now we are gonna show that $K(x)=y$:
Let $y^* \in Y^*$:
$$\langle y^*,y\rangle=lim_{j \to \infty}\langle y^*,K(x_{n_{i_j}})\rangle=\langle K^*y^*,x\rangle=\langle y^*,K(x)\rangle$$
where in the first step I used that the pairing $\langle \cdot,\cdot\rangle$ is continuous, in the second I use the property of the dual operator and the weak convergence and finally again the dual operator formula.
Since $$\langle y^*,y-K(x)\rangle = 0 \space \forall y^* \in Y^*$$
we can conclue by Hahn-Banach that  $y=K(x)$, hence $K(B)$ is closed. q.e.d.

I think you don't need the assertion that $K$ is compact in your proof, but that it is enough for $K$ to be bounded:  
You assume that $y := \lim_{n \rightarrow \infty} K(x_n)$ exists for a sequence $(x_n)_{n \in \mathbf{N}} \subset B$. Now if $X$ is reflexive, we may extract a subsequence $(x_{n_i})_{i \in \mathbf{N}}$ which converges weakly to $x \in B$. Then: $$\langle y^*, y \rangle = \lim_{i \rightarrow \infty} \langle y^*, K(x_{n_i}) \rangle = \lim_{i \rightarrow \infty} \langle K^* y^*, x_{n_i} \rangle = \langle y^*, K(x) \rangle$$ since $K^* y^* \in X^*$. Hence $y = K(x) \in K(B)$. 
I was thinking, whether compactness of $K$ could help you to prove the converse, but there's the following counterexample: Consider the non-reflexive space $\ell^1$ and the operator $K: \ell^1 \rightarrow \mathbf{R}$, $(x_n)_{n \in \mathbf{N}} \mapsto x_1$, which is compact, for it is of finite-rank; the image of the unit ball is $K(B) = [-1, 1]$.
