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.
A finite-dimensional subspace $U$ of $V$ is always closed: $U$ is complete with the induced norm (since all norms on $U$ are equivalent), so $\|u_n - x\| \to 0$ with $u_n \in U, x \in V$ implies that $u_n$ is a Cauchy sequence and hence converges to some $u \in U$. From this it is easy to see that $x = u \in U$ (because metric spaces are Hausdorff), i.e. $U$ is closed.
Infinite-dimensional subspaces may not be closed: Take for example $$c_{00}(\mathbb{C}) = \{(x_n)_{n \in \mathbb{N}} \in \mathbb{C}^{\mathbb{N}}: \exists\, m \in \mathbb{N} \text{ such that } x_n = 0 \, \forall {n \geq m}\}$$
and consider the inherited inner product of $\ell^2(\mathbb{C})$ on $V := c_{00}(\mathbb{C}) + \mathbb{C} \cdot z$ with $$z = (z_n)_{n \in \mathbb{N}} \in \ell^2(\mathbb{C}), \, z_n := \frac{1}{n}.$$
Then $V$ is not complete since it is not closed as a subset of $\ell^2(\mathbb{C})$. Furthermore $c_{00}(\mathbb{C})$ is dense in $\ell^2(\mathbb{C})$ and with this it is easy to see that $c_{00}(\mathbb{C})$ also is dense in $V$ with respect to the induced inner product. In particular, $c_{00}(\mathbb{C})$ is not closed in $V$ since $z \notin c_{00}(\mathbb{C})$.
Best Answer
I will show that for any subspace $W$, $W^{\perp\perp} = W^-$. (Closure)
Observe that $$ W^{\perp\perp} = \{v \in V\mid \forall w \in W^\perp, \langle v, w \rangle = 0\} = \{v \in V\mid \forall w \in V, (\forall u \in W, \langle u,w \rangle = 0 \implies \langle w,v \rangle = 0)\} $$ Then $u \parallel v$, so that $W \subseteq W^{\perp\perp}$. Since any orthogonal complement is closed, $W^{\perp\perp}$ is closed. Since $W^-$ is the smallest closed set containing $W$, we have $W^- \subseteq W^{\perp\perp}$.
Conversely, let $w \in W^{\perp\perp}$. Assume that $w \not\in W^-$. By the Hilbert Projection Theorem, ($W^-$ is convex) there exists $ w_0 \in W^-$ and some $w_1\mid w_1\neq 0, w_1 \perp W^-$ such that $w = w_0 + w_1$. Thus, $$ \langle w, w_1 \rangle = \langle w_0 + w_1, w_1 \rangle = \langle w_1, w_1 \rangle > 0 $$ hence $w \not\in W^{\perp\perp}$. Therefore $W^- \supseteq W^{\perp\perp}$.