Show reflexive normed vector space is a Banach space $X$ is a normed vector space. Assume $X$ is reflexive, then $X$ must be a Banach space.
I guess we only need to show any Cauchy sequence is convergent in $X$.
 A: Hint: (1) If $X$ is reflexive, $X$ is isomorphic to $X^{**}$. (2) Dual spaces are allways complete.

Regarding (2), we will prove, that $L(X,Y)$ the space of bounded linear operators from $X$ to $Y$ is complete in the operator norm if $Y$ is complete. Then (2) follows, as $X^* = L(X, \mathbb K)$ and $\mathbb K$ is complete. So let $(T_n)$ be an operator norm Cauchy sequence, then $(T_n x)$ is Cauchy for each $x$, as $\def\norm#1{\left\|#1\right\|}$
$$ \norm{T_nx-T_mx} \le \norm{T_n - T_m}\norm x $$
As $Y$ is complete, we may define $T\colon X \to Y$ by $Tx := \lim_n T_n x$. $T$ is linear, as the $T_n$ and the limit is, an bounded since 
$$ \norm{Tx} \le \sup_n\norm{T_n x} \le \sup_n\norm{T_n}\cdot \norm x $$
and Cauchy sequences are bounded. Now given $\epsilon > 0$, we can find a $N$, such that 
$$ \norm{T_n - T_m} < \epsilon, \text{  all $n,m \ge N$} $$
giving 
$$ \norm{T_n x - T_m x} < \epsilon, \text{   all $\norm x \le 1$, $n,m \ge N$} $$
for $m \to \infty$
$$ \norm{T_n x - T x} \le \epsilon, \text{   all $\norm x \le 1$, $n\ge N$} $$
that is $\norm{T_n - T} \le \epsilon$, $n \ge N$. So $T_n \to T$.
