Let $\beta=\{w_1,w_2,\ldots,w_k\}$ and $\gamma=\{x_1,x_2,\ldots,x_m\}$ be the bases for $W$ and $W^\perp$, respectively. It suffices to show that
$$\beta\cup\gamma=\{w_1,w_2,\ldots,w_k,x_1,x_2,\ldots,x_m\}$$
is a basis for $V$.
Given $v\in V$, then it is well-known that $v=v_1+v_2$ for some $v_1\in W$ and $v_2\in W^\perp$. Also because $\beta$ and $\gamma$ are bases for $W$ and $W^\perp$, respectively, there exist scalars
$a_1,a_2,\ldots,a_k,b_1,b_2,\ldots,b_m$ such that
$v_1=\displaystyle\sum_{i=1}^ka_iw_i$ and $v_2=\displaystyle\sum_{j=1}^mb_jx_j$. Therefore
$$v=v_1+v_2=\sum_{i=1}^ka_iw_i+\sum_{j=1}^mb_jx_j,$$
which follows that $\beta\cup\gamma$ generates $V$. Next, we show that
$\beta\cup\gamma$ is linearly independent. Given
$c_1,c_2,\ldots,c_k,d_1,d_2,\ldots,d_m$ such that
$\displaystyle\sum_{i=1}^kc_iw_i+\sum_{j=1}^md_jx_j={\it 0}$, then
$\displaystyle\sum_{i=1}^kc_iw_i=-\sum_{j=1}^md_jx_j$. It follows that
$$\sum_{i=1}^kc_iw_i\in W\cap W^\perp\quad\mbox{and}\quad
\sum_{j=1}^md_jx_j\in W\cap W^\perp.$$
But since $W\cap W^\perp=\{{\it 0}\,\}$ (gievn $x\in W\cap W^\perp$,
we have $\langle x,x\rangle=0$ and thus $x={\it 0}\,$), we have
$\displaystyle\sum_{i=1}^kc_iw_i=\sum_{j=1}^md_jx_j={\it 0}$. Therefore
$c_i=0$ and $d_j=0$ for each $i,j$ becasue $\beta$ and $\gamma$ are bases
for $W$ and $W^\perp$, respectively. Hence we conclude that $\beta\cup\gamma$ is linearly independent.
Best Answer
Let $M \subseteq \mathcal{H}$ be a linear subspace of the Hilbert space.
(i) Consider $\{\varphi_n\}$ as a Cauchy sequence in $M$
Because $\forall \varphi_n \in M \implies \forall \mu \in M^\perp: \langle\mu, \varphi_n\rangle=0\implies \varphi_n\in (M^\perp)^\perp$
Therefore, $M \subseteq M^{\perp\perp}$
(ii) $M^{\perp\perp}$ is an orthogonal complement and hence it is a closed linear subspace of $\mathcal{H}$.
Therefore, it is a sub Hilbert space. Hence complete. So $\{\varphi_n\}$ is a Cauchy sequence in $M^{\perp\perp}$ and hence it converges in $M^{\perp\perp}$:
$$\lim_{n \to \infty} \varphi_n= \varphi \in M^{\perp\perp}$$
We have shown that $\{\varphi_n\}$ as a Cauchy sequence in $M$ converges to $\varphi$ in $M^{\perp\perp}$. That is the topological closure of $M$ is $M^{\perp\perp}$:
$$\overline{M}=M^{\perp\perp}$$