See this question:Weak convergence and convergence of norms imply strong convergence in Hilbert space
But how can I find a counterexample if we relax the requirement that the space is Hilbert? I mean if we drop reflexivity of Banach spaces, then there should be a counterexample, but I can't think of any yet.
Take $c_0$, the space of null convergent sequences. Suppose $((x^m_n)_n)_m$ is a sequence in $c_0$ defined by $$x_n^m = \begin{cases} 1, & \text{if } n = 0 \text{ or } n = m + 1 \\ 0, & \text{otherwise.} \end{cases}$$ For example, the third point in the $c_0$ sequence, $(x_n^2)_n$, is the scalar sequence $1, 0, 0, 1, 0, 0, 0, \ldots$
Note that $\|(x_n^m)_n\| = 1$ for all $m$. I claim that the sequence converges weakly, but not strongly to $(x_n)_n$, defined by $$x_n = \begin{cases} 1, & \text{if } n = 0 \\ 0, & \text{otherwise.} \end{cases}$$ Note that $\|(x_n)_n\| = 1$ too, so convergence in norms is satisfied.
Recall that the dual space of $c_0$ can be considered to be $\ell^1$, where the sequence $(z_n)_n \in \ell^1$ acts upon $c_0$ by $$(y_n)_n \in c_0 \mapsto \sum_n z_n y_n.$$ So, let's fix $(z_n)_n \in \ell^1$. Then, applying this to $x^m - x$ as a functional on $c_0$, we get $$\sum_n z_n(x^m_n - x_n) = z_{m+1}.$$ But $z_{m+1} \to 0$ as $m \to \infty$, so $x^m$ converges weakly to $x$, as expected. Further, $\|x^m - x\| = 1$, achieving a maximum at $n = m + 1$, and so we do not have norm convergence.
An example in the Banach space $C([0,1])$. Let $$f_n(x)= \begin{cases}1- n x \textrm{ for } x \in [0, \frac{1}{n}] \\ nx -1 \textrm{ for } x \in [\frac{1}{n}, \frac{2}{n}] \\ 1 \textrm{ for } x \in [\frac{2}{n}, 1] \end{cases}$$ and $f(x) \equiv 1$. Then $f_n \overset{w}{\to} f$, $\|f_n\|= \|f\|=1$, but $\|f_n-f\|=1$ for all $n$.
Note: We have $f_n \overset{w}{\to} f$ if and only if $(\|f_n\|)_n$ is bounded, and $f_n(t) \to f(t)$ for all $t$.
$\bf{Added:}$ Another example in the same space $C([0,1])$, the functions are polynomial: $f_n(x) = 1 - 4 x^n(1-x^n) \overset{w}{\to} f(x) \equiv 1$, $\|f_n\|=\|f\|=1$, and $\|f_n- f\| = |f_n( {\frac{1}{\sqrt[n]{2}}})-1|=1$.
It seems that if $X$ is a reflexive Banach space, $x_n \overset{w}{\to} x$, and $\|x_n\|\to \|x\|$, then $x_n \to x$ ( in the norm topology). I have seen a proof of this for $L^p$ spaces, perhaps worth understanding.
Note that if $X=\ell^1$ ( absolutely summable sequences), and $x_n \overset{w}{\to} x$, then $x_n \to x$ ( Schur's theorem). Note that $\ell^1$ is not reflexive.
Here is an example in $C([0,1])$. Define $$ f_n(x) = x + \max\left(0, 1-n |x-\frac1n| \right). $$ So $f_n$ is a continuous function plus a moving hat of shrinking diameter. Then $f_n \rightharpoonup f$ and $\|f_n\|\to \|f\|$ for $f(x)=x$.
