Counterexample $(V^\perp )^\perp = V$ when $V$ is not a closed subspace Let $(X, \langle\cdot, \cdot \rangle)$ be a Hilbert space with inner product $\langle\cdot, \cdot \rangle$ and induced norm $\rVert \cdot \lVert$.
We know that if $V\subseteq X$ is a closed subspace, then  $(V^\perp  )^\perp = V$, where $V^\perp$ denotes the orthogonal complement with respect to the inner product.  
Now let $X:= L^2([0,1], \mathcal L^1, \lambda^1)$ be the set of Lebesgue square integrable functions on $[0,1]$.
Define $V:= \{\varphi \in X: \varphi(0) = 0 \}$. Clearly, $V$ is a subspace of $X$.
Now I showed that $V \neq(V^\perp  )^\perp$, i.e. we have a strict inclusion, by explicitly constructing a function that is in $(V^\perp  )^\perp$, but not in $V$.
To understand this better I tried to show that $V$ is not closed with respect to the $L^2$-norm. So I tried to construct a sequence $(\varphi_n)_{n\in \mathbb N}$ such that $\varphi_n(0) = 0$ for all $n\in \mathbb N$ but $\varphi_n \to \varphi$ with $\varphi(0) \neq 0$. However this was kind of tedious dealing with the square in the integrand. Is there an easy way to construct a sequence or some other, better/cleaner option to disprove closedness?
 A: For any subset $S$ of a inner product space, $S^\perp$ is closed(use continuity of the inner product). So $V^{\perp\perp}$ never equals to $V$ when $V$ is not closed.
A: To get a complete inner product space from Lebesgue integrals, you must consider equivalence classes of functions defined up to sets of measure zero. Consequently predicates like $\phi(0)=0$ do not make sense for elements of $L^2([0,1])$.
We could instead ask the question about a space where evaluation does make sense, like the continuous functions $C([0,1])$ with uniform convergence or the space of pointwise convergence $\mathbb{R}^{[0,1]}$. Note that the space of continuous functions is not complete in the $L^2$ norm, so this is no longer a Hilbert space.
In both cases, the evaluation function (a function which takes a function and returns its value at a fixed input) $\text{ev}_0$ is continuous, hence the set $V=\{\phi\colon \phi(0)=0\}$ is the preimage of a point under a continuous function $V=\text{ev}_0^{-1}(0)$, hence it is closed. You cannot show this set is not closed, because it is in fact closed.
The example sequence given by Ice sea does not show that the set is not closed in $\mathbb{R}^{[0,1]}$, because it converges pointwise to a function with $\phi(0)=0$. It doesn't show that the set is not closed in $C([0,1])$, because it does not converge uniformly. And it does not show that the set is not closed in $L^2([0,1])$ because the set is not defined there.
For an example of a set that is not closed, just think of things which include only finite combinations, but not limits. For example, the space of polynomial functions $P([0,1])$ makes sense as a subset in any of these three function spaces. $e^x$ is the limit of a sequence of polynomials. Hence $e^x$ is in $(P([0,1])^\perp)^\perp$ but not $P([0,1])$.
Or consider the function space $L^2([-1,1])$. This space can be split into even functions $E=\{\phi\colon \phi(x) = \phi(-x)\}$ and odd functions $O=\{\phi\colon \phi(x) = -\phi(-x)\}$. The both subspaces are closed. A basis for the even functions is $\cos(n\pi x)$. The function $\left|1-x\right|$ is even. It's perpendicular to all odd functions. It's in $(\langle\cos(n\pi x)\rangle^\perp)^\perp,$ but not $\langle\cos(n\pi x)\rangle,$ since it's not a finite linear combination of cosine functions.
A: First, if the element in your space is considered to be an "equivalent classes", then $V=L^2[0,1]$. Obviously, it is closed. To make you problem well-defined, we consider $L^2[0,1]$ to be space of functions instead of equivalent classes. 
It is easy to disprove the closendess  in this case.
 $f_n(x)=(1-x)^n-1$ for all $x\in[0,1]$. Then $f_n(x)\to f(x)=-1$ in $L^2[0,1]$  as $n\to\infty$. In addition, $f_n(0)=0$.  To see the convergence, we only need the dominated convergence theorem. First $f_n(x)\to f(x)$ a.e. and they are are both bounded from above. So $f_n\to f$ in $L^2[0,1]$. i.e., 
$$
\int |f_n(x)-f(x)|^2 dx \to 0
$$ 
A: Take the “prototype” of all Hilbert non finitely generated but separable space, namely $l^2$, the space of (complex) sequences $(a_n)$ such that
$$
\sum_{n\ge0}|a_n|^2
$$
converges (for a real Hilbert space just use real sequences). The Hilbert form is
$$
\langle (a_n),(b_n)\rangle=\sum_{n\ge0}\overline{a_n}b_n
$$
(switch the conjugation if you use the opposite convention about linearity).
The subspace we want to consider is $V$ consisting of the “eventually null” sequences, that is all $(a_n)$ such that there exists $N$ with $a_n=0$ for every $n\ge N$.
It's easy to prove that $V$ is dense in $l^2$. Thus $V^{\perp\perp}=l^2$. Indeed, given $s=(a_n)\in l^2$, define
$$
s^{(m)}(n)=\begin{cases}
a_n & n\le m\\
0 & n>m
\end{cases}
$$
Then $s^{(m)}\in V$ and $\lim_{m\to\infty}s^{(m)}=s$ (check it), so the statement about density follows. Since $V$ is dense, $V^{\perp}=\{0\}$.
