In a finite dimensional inner product vector space, yes, because for every subspace $W$ we have $(W^{\perp})^{\perp} = W$. In particular, if $W\neq V$, then $W^{\perp}\neq\mathbf{0}$, since $\mathbf{0}^{\perp}=V$. That $W$ is a sum is immaterial.
In the infinite dimensional case, no. You can have a proper subspace whose orthogonal complement is trivial. E.g., in the vector space of all square summable real sequences, viewed as a subspace of $\mathbb{R}^{\infty}$, the span of the basis vectors $\mathbf{e}_i$ is a proper subspace that is dense, so its orthogonal complement is trivial. The same holds in any infinite dimensional Hilbert space, by taking the span of a Hilbert basis.
Added. If your space does not have an inner product, then the very concept of "orthogonality" has no meaning, so the answer is Mu.