Let $(X,\|\cdot\|)$ be a separable, infinite dimensional Banach space with a Hamel basis $\{v_i,i\in I\}$ and $\|v_i\|=1$ for all $i\in I$. Define $\langle\cdot \rangle:X\to [0,\infty]$ by $\langle x\rangle = \sum_{i} |\lambda_i|$ where $x=\sum_{i} \lambda_i v_i$. Check that this is a norm and hence show that closed graph theorem fails if codomain is not complete.
Checking that the function is norm is easy.
I suspect the map I should be considering is the identity $id_X:(X,\|\cdot\|) \to (X,\langle \cdot \rangle)$.
The graph of this map is closed: say $\|x_n-x\|\to 0$ and $\langle x_n-y\rangle \to 0$. The condition $\langle x_n-y\rangle \to 0$ forces "point-wise" convergence of $x_n$ to $y$ in the sense of given hamel basis (i.e. the coordinates of $x_n$ must converge to that of $y$).
If $x_n=\sum_{i} \lambda_{i}^{n} v_i$ and $y=\sum_i \mu_i v_i$
$$ \|x_n-y\|=\left\|\sum_i (\lambda_i^n - \mu_i)v_i \right\|\le \sum_i |\lambda_i^n - \mu_i| \|v_i\| \to 0 $$
because $\lambda_i^n \to \mu_i$ (and above probably also shows that $\|x\|\le \langle x\rangle$ for all $x$). So $y=x$.
I still need to show: this map is unbounded, and that $X$ with $\langle\cdot\rangle$ is not complete, but I'm stuck here.
Edit: On a second thought, above argument does not work because convergence of each term in the series does not guarantee convergence of series to $0$. For instance $\sum_n 1/n = \infty$ but $1/n\to 0$... so that leaves me in complete darkness
Start with the map in the other direction, $S = \operatorname{id}_X \colon (X, \langle\,\cdot\,\rangle) \to (X, \lVert\,\cdot\,\rVert)$. For $x = \sum_i \lambda_i v_i$, we have
$$\lVert Sx\rVert = \Biggl\lVert \sum_{i \in I} \lambda_i v_i\Biggr\rVert \leqslant \sum_{i\in I} \lvert \lambda_i\rvert\cdot \lVert v_i\rVert = \sum_{i\in I} \lvert \lambda_i\rvert = \langle x\rangle.$$
Thus $S$ is continuous (with $\lVert S\rVert = 1$ - equality since $\lVert v_i\rVert = \langle v_i\rangle$), and hence its graph is closed. The graph of $T = S^{-1}$ is therefore also closed - the map $(x,y) \mapsto (y,x)$ is a homeomorphism between $(X,\langle\,\cdot\,\rangle)\times (X,\lVert\,\cdot\,\rVert)$ and $(X,\lVert\,\cdot\,\rVert)\times (X,\langle\,\cdot\,\rangle)$.
Now, by the open mapping theorem, or the closed graph theorem, it is equivalent to show that $T$ is unbounded or that $(X,\langle\,\cdot\,\rangle)$ is not a Banach space. I think it is easier to show the latter.
Choose a countably infinite subset $J \subset I$ and enumerate it. For simplicity, write $w_n = v_{j_n}$. Then consider
$$x_n = \sum_{k = 0}^n 2^{-k} w_k$$
for $n \in \mathbb{N}$. Verify that $(x_n)_{n\in \mathbb{N}}$ is a Cauchy sequence in $(X,\langle\,\cdot\,\rangle)$. Show that this sequence doesn't converge with respect to $\langle\,\cdot\,\rangle$, however.
Note: the separability of $X$ (with respect to $\lVert\,\cdot\,\rVert$) plays no role in this proof.
Since the separability of $(X,\lVert\,\cdot\,\rVert)$ was given as a hypothesis, the expected proof (that the two norms aren't equivalent) was probably using the fact that $(X,\langle\,\cdot\,\rangle)$ is not separable:
The Hamel basis $\{ v_i : i \in I\}$ is an uncountable discrete subset of $(X,\langle\,\cdot\,\rangle)$ - $\langle v_i - v_j\rangle = 2$ for $i \neq j$ - and a separable metric space is second countable, hence a discrete subset of a separable metric space can be at most countable.