Def 1: Let $X$ and $Y$ be metric spaces and $T:X\to Y$, then $T$ is closed if $x_n \to x$ in X and $T(x_n) \to y$ in $Y$ implies $y=T(x)$.
Def 2: Let $X$ and $Y$ be metric spaces and $T:X\to Y$, then $T$ is closed if for every closed set $E\subset X$, $T(E)$ is closed.
Definition 1 was covered in my Functional Analysis course, while definition 2 was done in Topology. These two definitions are not equivalent. Continuous maps like $T(x)=\frac{1}{1+x^2}$, where $x \in \mathbb{R}$, are not closed by definition 2, but are closed by definition 1. For $f : \mathbb{R} \to \mathbb{R}$ defined by $f(x) = 1$ if $x<1/2$ and $0$ otherwise, is closed by definition 2 because the image is a subset of $\{0,1\}$ for any closed subset of $\mathbb{R}$, however, for the sequence $x_n = \frac{1}{2}-\frac{1}{n}$, $x_n \to 1/2$ and $f(x_n) \to 1$, but $f(1/2) =0\ne 1$.
Can we put further restrictions on $T$ so that the definitions become equivalent? like $T$ being a linear operator between norm spaces.
Definition 1 is equivalent to the condition that the graph of $f: \{(x,f(x))\mid x\in X\}$ is closed in the topological space $X\times Y$ (when $X$ and $Y$ are metric spaces, or more generally first-countable spaces). This condition is (in general) totally unrelated to the condition that $f$ is a closed map (Definition 2). It seems that some authors refer to a function with closed graph as a closed map, but this terminology seems very likely to cause confusion.
In many situations, having a closed graph (Definition 1) is equivalent to being continuous. See the Wikipedia page Closed graph theorem. As you pointed out in your question, the conditions "$f$ is continuous" and "$f$ is a closed map" are orthogonal.