Counterexample to $\mathcal{H} = M \oplus M^{\perp}$ where $M$ not closed, $\mathcal{H}$ Hilbert

Question

Let $\DeclareMathOperator{\H}{\mathcal{H}} \H$ be an Hilbert space and $M \subset \H$ a closed subspace. Then $\H = M \oplus M^{\perp}$, where $\oplus$ denotes the orthogonal direct sum and $^\perp$ the orthogonal complement.

Since we require $M$ to be closed and it is essential for the proof, I suspect there's a counterexample to this statement for non-closed $M$ but haven't been able to come up with one. Any hints are appreciated.

Answer 1

Let $M$ be any dense proper subspace. Then $M^{\perp}=\{0\}$.

For an explicit example take $M$ to be the set of sequences with at most finite number of non-zero terms in $H=\ell^{2}$.

Answer 2

Take any non-closed $M \subset \mathcal H$. Then, every $x \in \bar M \setminus M$ is not contained in $M + M^\perp$.