Let $c$ be the vector space of all real convergent sequences, with the norm
$$\|(a_n)_{n\in \mathbb N}\|_\infty = \sup\{|a_n|:n\in \mathbb N\}$$
Let $c_0$ be the subspace of $c$ of sequences with limit $0$, with the standard $\sup$-norm.
Define $T:c \to c_0$ by
$$T((a_n))=(a,a_1-a,a_2-a,\dots)$$
where $a=\lim_{n \to \infty}a_n$.
Question: How do I prove that $T$ is a bijection (1-1 and onto), and also continuous?
Presumably you are using $\mathbb N = \{1,2,3,\dots\}$.
The first thing is to prove that the function $(l,a_1-l,a_2-l,\dots)$ actually converges to $0$. This is easy.
Now prove it is injective. suppose $f((a_n))=f((b_n))$. Then we must have that both limits $l_a$ and $l_b$ are equal, we also have $l_a-a_k=l_b-b_k$ for all $j\in \mathbb N$. Since $l_a=l_b$ we have $a_k=b_k$ for all $k$. So the two sequences are equal.
To prove it is surjective. suppose you have a sequence $(a_1,a_2,\dots )$ that converges to $0$. Then the sequence $(a_1+a_2,a_1+a_3,a_1+a_4,\dots)$ converges to $a_1$ (because $a_n$ converges to $0$).
It follows that $f(a_1+a_2,a_1+a_3,\dots)=(a_1,a_2,a_3,\dots)$. So it is surjective.