Use the closed graph theorem: if $x_n\to x$ and $Tx_n\to y$, hence by surjectivity $y=Tu$ for some $u\in X$. We conclude by the assumption that $\lVert x_n-u\rVert\to 0$ hence $u=x$ and $y=Tx$, proving that the graph is closed.
Given two structures of $A,B$ of some kind, you first have to guess what a structure-preserving 'map', i.e. a (homo-) morphism $A \to B$ (from $A$ to $B$) is. For the sake of simplicity, we are going to assume, that a morphism is a function with some special properties (this need not be the case, but it very often is). If you have this, then $A$ is commonly called isomorphic to $B$, if there is an isomorphism $A\to B$.
An isomorphism $f : A\to B$ is a morphism $A\to B$ with an inverse, that is a morphism $g : B\to A$ with $g\circ f = \operatorname{id}_A$ and $f\circ g = \operatorname{id}_B$. Of course, in our case we then have $g = f^{-1}$. So, being an isomorphism implies being bijective. The converse is not generally true, because even if $f$ is a bijective morphism, $f^{-1}$ is not necessarily a morphism too (this is indeed the case for topological spaces and their morphisms), but it is true for example for linear maps.
How do you decide what a morphism should be, i.e. what kind of special properties a map between structures should have? Well, it depends. Sometimes its obvious (vector spaces), sometimes it is not (metric spaces). But in any case, the morphisms should preserve "all" the structure. So vector space morphisms (linear maps) should preserve addition and scalar multiplication and a metric space morphism should preserve the metric.
Unfortunately or maybe interestingly, there are many things you could consider "preserving the metric". One kind of possible morphisms are simply called "distance-preserving", but there are other possibilities (short maps, lipschitz-continuous maps, etc.). Because in this case, we have many notions of morphisms, someone decided to use the term "isometric" in place of "isomorphic".
The general theory dealing with "morphisms between structures" is called "category theory".
Best Answer
Take any infinite-dimensional Banach space and choose its Hamel basis $\left(e_i\right)_{i\in I}$, say, with $\left\|e_i\right\|=1$. Define $f\colon X \to X$ by
$$ f(e_i) = \alpha_i e_i $$
where ${\alpha_i}$ is any unbounded collection of (nonzero) numbers. Clearly $f$ is a bijection, and $f$ is unbounded, hence discontinuous.
Now, the above relies critically on the Axiom of Choice. If I recall correctly, there was a similar question here, and it's been pointed out that in certain models with negation of Axiom of Choice there are no discontinuous linear maps from Banach spaces to normed linear spaces at all. So there can be no "truly constructive" example.
It's probably a bit off-topic, but such natural examples exist for incomplete normed linear spaces. A well-known one is a derivative operator for $C^\infty\left([0,1]\right)$ with supremum norm. It's not injective, but it is on a quotient by subspace of constant functions.