In one of my functional analysis tests I was asked to prove the following theorem:
A Banach space is reflexive and has the Schur property (every weakly convergent sequence is convergent in the norm), iff it's finite dimensional.
I've been thinking about using the fact that a space is finite dimensional iff the weak and strong topologies coincide, but I'm not sure if that helps in any way. i also thought about using the fact that X will be isomorphic to X**.
The suggestion made in the comment works (and is the only solution to this problem that I am aware of). However it only uses the easy direction of Eberlein-Smulian (weakly compact sets are weakly sequentially compact) so maybe the intended solution is just to provide a proof of that easy direction. In what follows I fix a Banach space $X$
Proof: Let $K \subseteq X$ be weakly compact and let $(x_n)$ be a sequence in $K$. Define $\tilde{X} = \overline{\operatorname{span}\{x_n: n \geq 1\}}$. Then $\tilde{X}$ is separable and the weak topology on $\tilde{X}$ coincides with the restriction of the weak topology on $X$ to $\tilde{X}$. Hence is suffices to see that $(x_n)$ has a weakly convergent subsequence in $\tilde{X}$. This is straightforward since $x_n \in \tilde{K} = \tilde{X} \cap K$ for every $n$ and $\tilde{K}$ is weakly compact and separable and in particular, the weak topology on $\tilde{K}$ is metrizable. For metrizable spaces, it is a basic fact that the notions of compactness and sequential compactness agree, so that $(x_n)$ has a convergent subsequence in $\tilde{K}$ and hence in $K$ also.
The desired result then follows since if $X$ is reflexive, $B_X$ is weakly compact and hence weakly sequentially compact. Then if $X$ also has the schur property it follows that $B_X$ is sequentially compact for the norm topology. This is not the case if $X$ is infinite dimensional so that $\dim X < \infty$.