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Given a Banach space $X$ which can be written a direct sum of two subspaces $Y\oplus Z$ and the $u\in B(X), w\in B(Y), v\in B(Z)$ and $u=w\oplus v$. where $B(\cdot)$ denotes the space of bounded operators on the space.

So is it true that $u$ is compact iff both $v$ and $w$ are? If not, what would be the counter-example?

The $\Rightarrow$ is indeed true, as seen in Direct sum of compact operators is compact. Nonetheless, the reverse direction requires that every compact operators to be the norm-limit of finite-rank operators, which is not the case in general Banach spaces.

P.S. If $X, Y, Z$ are all Hilbert spaces, this is indeed true. The same link above addresses the question.

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The other implication is true as well, at least if you assume that $Y$ and $Z$ are closed in $X$. In this case, the projection maps are bounded (by the closed graph theorem) which is needed in the proof.

Start with a bounded sequence $(x_n)=(y_n\oplus z_n)$ in $X=Y\oplus Z$. You are looking for a subsequence $(x_{n_k})$ such that the sequence $(u(x_{n_k}))$ is convergent.

The sequences $(x_n)$ and $(z_n)$ are bounded in $Y$ and $Z$ by boundedness of the projection maps. Now, use the compactness of $v$ and $w$... (You will certainly be able to finish).

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