Let $T: c\to c$ be a bounded linear operator, where $c$ is the vector space of convergent real sequences. How can we prove that there exists an infinite matrix $A=(a_{n,k}: n,k\ge 1)$ such that $T(x)=Ax$ for all $x \in c$?
The proof is very easy replacing $c$ in both places with $c_0=\{x \in c: \lim_n x_n=0\}$ (indeed, in such case, $\{e^i: i\ge 1\}$ is a Schauder basis for $c_0$, where $e^i \in c$ is the sequence which is constantly equal to $0$, except $e_i^i=1$).
It depends on what you mean by "$Ax$".
If you consider the standard definition ma matrix multiplication (ie. with respect to the "basis" $\{e^i: i\ge 1\}$) it does not hold. Consider for example the operator $T$ defined by $$T(x) = (\lim x_n, 0, 0 ,\dots).$$ Then $T:c \rightarrow c$ is a bounded linear operator but cannot be expressed by a matrix multiplication as the limit does not depend on any finite initial segment of a sequence. In a certain way, this is the only operator on $c$ which can't be expressed by matrix multiplication, see below.
However, the space $c$ can be identified with $c_0 \oplus \mathbb{C}$ via identification $(x_n) \leftrightarrow ((x_n - \lim x_k), \lim x_k)$. Hence it has a Schauder basis $(u, e^1,e^2, \dots)$, where $u = (1,1,1,\dots)$, and every bounded linear operator $T:c \rightarrow c$ can be expressed by matrix multiplication with respect to this basis.