We have $\dim W$ since $(v_1, v_2, v_3)$ is linearly independent. To see that, suppose $$c_1v_1 + c_2v_2 + c_3v_3 = 0$$ for some $c_1,c_2,c_2\in\mathbb R$. This implies
\begin{align}
c_1 + c_2 + c_3 &= 0\tag 1\\
c_1 + c_3 &= 0\tag 2\\
c_1 + c_2 &= 0\tag 3\\
c_1 &= 0\tag 4
\end{align}
$(4)$ directly implies $c_1=0$, which in turn implies $c_2=c_3 = 0$ from $(2)$ and $(3)$.
To construct an orthogonal basis for $W$, there is a standard inductive algorithm called Gram-Schmidt. In general, suppose we have a basis $(v_1, \ldots, v_n)$ for a subspace $V$ of $\mathbb R^m$ where $n<m$ (the case where $n=m$ isn't interesting because we can just take the standard basis). The algorithm proceeds as follows:
- Let $V_1 = \operatorname{Span}(\{u_1\})$, where $u_1=v_1$.
- Given $V_i$, $i\leqslant 1\leqslant n-1$, set $V_{i+1} = V_i \cup \{u_{i+1}\}$ where $$u_{i+1} = v_{i+1} - P_{V_i}(v_{i+1}),$$
where $P_{\cdot}$ denotes the projection operator, i.e.
$$P_{V_i}(v_{i+1}) = \sum_{j=1}^i \frac{\langle v_{i+1}, v_j \rangle}{\langle v_j, v_j \rangle} v_j. $$
In this example, we have $u_1=v_1=(1,1,1,1)$, and then
\begin{align}
u_2 &= v_2 - P_{V_1}(v_2) = (1,0,1,0) - \frac{\langle (1,0,1,0), (1,1,1,1) \rangle}{\langle (1,1,1,1), (1,1,1,1) \rangle} \\
&= (1,0,1,0) - \frac24(1,1,1,1)\\
&= \left(\frac12,-\frac12,\frac12,-\frac12\right).
\end{align}
Finally,
\begin{align}
u_3 =&\ v_3 - P_{V_2}(v_2)\\
=&\ (1,1,0,0)\\ &- \left[\frac{\langle (1,1,0,0), (1,1,1,1) \rangle}{\langle (1,1,1,1), (1,1,1,1) \rangle}(1,1,1,1)+\right.\\ &\quad\quad\left.\frac{\langle (1,1,0,0), \left(\frac12,-\frac12,\frac12,-\frac12\right) \rangle}{\langle \left(\frac12,-\frac12,\frac12,-\frac12\right), \left(\frac12,-\frac12,\frac12,-\frac12\right) \rangle} \left(\frac12,-\frac12,\frac12,-\frac12\right) \right]\\
=&\ (1,1,0,0) - \left(\frac24 (1,1,1,1) + 0 \right)\\
=&\ \left(\frac12, \frac12, -\frac12, -\frac12\right).
\end{align}
(Notice that conveniently $v_3\perp u_2$, saving some computation). Hence, our orthogonal basis is
$$\left(u_1, u_2, u_3 \right) = \left(\left(1,1,1,1\right), \left(\frac12,-\frac12,\frac12,-\frac12\right),\left(\frac12, \frac12, -\frac12, -\frac12\right) \right)$$
To convert this orthogonal basis, we need only divide the basis elements by their norm, i.e. compute $e_i = \frac1{\|u_i\|}$ where $\|u_i\| = \langle u_i,u_i\rangle^{\frac12}$. I'll spare the computation, but the result is
$$(e_1, e_2, e_3) = \left(\left(\frac14, \frac14, \frac14, \frac14 \right),\left(\frac12,-\frac12,\frac12,-\frac12\right),\left(\frac12, \frac12, -\frac12, -\frac12\right) \right). $$
Best Answer
No, it is false, it needs further assumptions, also as stated in the comments, be aware that we are talking subspaces here, not vectors! for example (as gandalf61 said) take $Z=W\subset V$ then we have, since $\langle\_,\_ \rangle$ is non degenerate, for all $0 \neq z \in Z$ a $w \in W=Z$ such that $\langle z , w \rangle \neq 0$ in particular (to give a very concrete example) you could pick $w=z$ and then we already know that $$\langle z, w\rangle = \langle z, z\rangle = 0 \iff z=0$$
Also, be aware that only the concrete counterexample with $z=w$ needs an inner product, for the first one it suffices to have bilinear form $\langle \_ ,\_ \rangle$ inducing a non degenerate bilinearform on $Z=W$.