Orthogonal completion in nonhilbert spaces Let $X$ be some Hilbert space. There is the widely known theorem in functional analysis which states that for each closed subspace $H\subset X$ we have $H\bigoplus H^{\perp}=X$.
Now we do not suppose $X$ to be Hilbert space (but it is endowed with a scalar product). Is it true that there always exists such closed subspace $H\subset X$ such that $H\bigoplus H^{\perp}\ne X$?

So, I know that the completness of $X$ is important in the proves given in books, but how to build such closed subspace $H$?
 A: The meaningful question is: Does every closed subspace of a Banach space X have a topologically complementary subspace. The answer to this is: if and only if X is Hilbert. 
Edit: I misunderstood the original question. Examples of closed hyperplanes in pre-Hilbert spaces which do not have orthogonal complement are constructed here:
A counterexample to theorem about orthogonal projection. The point is that if a hyperplane $H$ in $X$ has an orthogonal complement $H^\perp=<u>$ ($u$ is the unit vector), then for each vector $v\in X$ one can define the orthogonal projection to $H$ by the usual formula
$$
p(v)= v- (u\cdot v)u. 
$$
However, in the linked post one finds examples of closed hyperplanes in pre-Hilbert spaces $X$ so that for some $v\in X$, the orthogonal projection to $H$ does not exist.  
A: Maybe something like this works? 
Starting with $\ell^2(\mathbb{R})$, let $e_i$ be the sequence with a $1$ in the $i$th position, and $0$ elsewhere. Then take $X$ to be the linear span of the $e_i$'s and some sequence $v$, where $v$ has infinitely-many non-zero entries - i.e. the subspace of all (square-summable) sequences with only finitely-many non-zero entries and some sequence like $\left(\dfrac{1}{n}\right)$. Then $X$ is a real inner-product space, but it isn't complete.
Then, if we let $H = \textrm{span}( e_2,\dots,e_n,\dots)$ (i.e. we have left out $e_1$ and $v$), then the orthogonal complement is $\textrm{span}(e_1)$, and $v$ is not in $H$ or $H^\perp$.
It is quite possible I have made a mistake in this answer, but hopefully there is something worthwhile here.
