When is the image of a linear operator (between Banach spaces) closed?

A case in which it is true that $R(T)$ is closed is if $T$ is bounded from below, i.e., there exists $c > 0$ such that $\|Tx\| \geq c\|x\|$ for all $x \in X$. To prove this, notice first that since $$Tx = 0 \implies 0 = \|Tx\| \geq c\|x\| \geq 0 \implies x= 0$$ $T$ must be injective. Now let $x_n \in R(T)$ be such that $x_n \to x$. Then we have that $$ \|x_n - x_m\| = \|TT^{-1}x_n - TT^{-1}x_m\| \geq c\|T^{-1}x_n - T^{-1}x_m\|$$ so that $T^{-1}x_n$ is a Cauchy sequence which must converge to some $z \in X$. But we have $$T(z) = T(\lim_{n\to\infty} T^{-1}x_n) = \lim_{n\to\infty} TT^{-1}x_n = \lim_{n\to\infty} x_n = x$$ which means that $x \in R(T)$ and thus $R(T)$ is closed.

In fact, if $T$ is assumed injective then this is a characterization of $T$ having closed range. For if $R(T)$ is closed in $Y$ then $R(T)$ is a Banach space under $\|\cdot\|_Y$ in its own right and the open mapping theorem shows that $T^{-1}:R(T) \to X$ is continuous which means there exists $C > 0$ such that $$\|T^{-1}x\| \le C\|x\|$$ or $$\|Tx\| \geq \frac{1}{C}\|x\|$$ for all $x \in X$.

Consider $X=Y:=\ell^\infty$, the normed space of bounded sequences. Let $T(x)(n):=\frac{x(n)}{n^2}$; this gives a linear continuous operator. Let $y\in T(X)$, then $\{n^2y(n)\}$ is bounded, and the converse holds. So $$T(X)=\{y,\sup_n|n^2y(n)|<\infty\}.$$ Let $x^{(n)}:=\sum_{j=1}^nj^{-1}e(j)$; it converges in $\ell^\infty$ to the sequence $x=\{n^{-1}\}$. Furthermore, $x^{(n)}\in T(X)$. But $x\notin T(X)$.