Let $X$ be a Banach space and $M$ be a subspace of $X$. Then we know $X= M\oplus N$ where $N$ is a subspace of $X$. Let $T:X \to X$ be a bounded linear operator.
I want to prove that
$ T= \begin{pmatrix} A&B\\ C&D\\ \end{pmatrix} $
where
$A:M \to M$
$B: M \to N$
$C:N \to M$
$D:N \to N$
are bounded linear operators.
This question is ambiguous. The matrix notation makes no sense for an operator in a general space unless you define how to evaluate that. I will provide such an interpretation, and prove that the statement is true in Hilbert spaces. We start with the definition.
$\textbf{Definition:}$ Assume that
$$X= M \oplus N,$$ where $X$ is a Hilbert space and $M$ and $N$ are closed subspaces of $X.$ Let $A:M\to M,\; B: N\to M,\; C: M\to N, \; D: N\to N$ be bounded linear operators. The operator $T$, denoted by
$$T=\begin{bmatrix} A &B \\ C & D \\ \end{bmatrix} $$ is defined as $$Tz= \left(A(P^M z)+B(P^N z)\right)+ \left(C(P^M z)+D(P^N z)\right),$$ where $P^M$ and $P^N$ denote the projection operators on $M$ and $N$ respectively, that is, if $z= x+y,\; x\in M,\; y\in N,$ then
$$P^Mz=x,\; P^N z=y.$$ Note that this definition is intrinsically related to the matrix representation, since $z= P^Mz+P^Nz,$ so that you can 'represent' $z$ as $$z= \begin{bmatrix} P^M z \\ P^N z\\ \end{bmatrix}$$ and we have the analogy with matrix multiplication.
Now, it is well known that in Hilbert spaces, if $S$ is a closed subspace, then the projection operator $P^S$ is continuous. With this in mind, we prove the following:
$\textbf{Statement:}$ If $X$ is a Hilbert space and $X=M\oplus N,$ with $M$ and $N$ closed subspaces, and $T:X\to X$ a continuous linear operator. Then $T$ admits a matrix representation of the form $$T=\begin{bmatrix} A &B \\ C & D \\ \end{bmatrix}. $$ $\textbf{Proof:}$ Put
$$A= (P^M \circ T)|_M,\;B= (P^M \circ T)|_N,\;C= (P^N \circ T)|_M,\;D= (P^N \circ T)|_N.$$ Here $U|_S$ denotes the restriction of the operator $U$ to the subspace $S.$ By construction, these operators are continuous. Consider now the operator $$T' = \begin{bmatrix} A &B \\ C & D \\ \end{bmatrix}.$$ We claim that $T=T'.$ In order to seee this, choose any $z \in X.$ Then
$$T'z = \left(A(P^M z)+B(P^N z)\right)+ \left(C(P^M z)+D(P^N z)\right)= $$ $$\left(P^M(T(P^M z))+P^M(T(P^N z))\right)+ \left(P^N(T(P^M z))+P^N(T(P^N z))\right)$$
$$= P^M(T(P^M z +P^N z ) ) +P^N(T(P^M z +P^N z ) )$$
$$= P^M(Tz ) +P^N(Tz ) = Tz,$$ as desired.
