Suppose that $X$ and $Y$ are two normed spaces over the same field ($\mathbb{R}$ or $\mathbb{C}$). Show that the range of a bounded linear operator $T \colon X \to Y$ need not be closed in $Y$.
Kreyszig gives the following hint:
Consider the operator $T \colon \ell^\infty \to \ell^\infty$ defined by $$Tx := y = (\eta_j), \, \mbox{ where } \, \eta_j := \frac{\xi_j}{j} \, \mbox{ for all } x := (\xi_j) \in \ell^\infty.$$
How do we characterise the range of this operator and show that the range is not closed in $\ell^\infty$?
Let $e_n$ be the element of $\ell_\infty$ whose $m$'th coordinate is $1$ if $m=n$ and $0$ otherwise. The closed linear span of $\{e_n\mid n\in \Bbb N\}$ in $\ell_\infty$ is the space $c_0$ of sequences that tend to $0$.
For your operator, we have $T(je_j)=e_j$; so the range of $T$ contains each $e_j$. Since the range of a linear operator is a linear space, it follows that the range of $T$ contains the linear span of the $e_j$. So, if the range of $T$ is closed, it must contain the space $c_0$ (in fact it would be equal to $c_0$).
But this is not the case: the vector $y=(1/\sqrt j)$ is in $c_0$ but is not on the range of $T$. If $Tx=y$, then the $j$'th coordinate of $x$ would have to be $\sqrt j$. The sequence $(\sqrt j)\notin\ell_\infty$.