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 $W^- \supseteq W^{\perp\perp}$.
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
If $S$ is not a subspace of $V$, then $\dim S$ does not make sense. You already proved that for any subset $S$, ${\rm span}(S) \subseteq (S^\perp)^{\perp}$, ok. Now, observe that $$\dim {\rm span}(S)^\perp = \dim S^\perp \quad \mbox{and}\quad \dim {\rm span}(S) + \dim S^\perp = \dim V < +\infty.$$The second formula follows from the rank-nullity theorem applied to $V \ni v \mapsto \langle v, \cdot\rangle|_{{\rm span}(S)} \in {\rm span}(S)^*$, as having finite-dimension implies that $\dim {\rm span}(S)^* = \dim {\rm span}(S)$.
Then you know that ${\rm span}(S) \subseteq (S^\perp)^\perp$ and both of them have the same finite dimension. Then they have to be equal.