Open linear subspace of a Hilbert space.

If $M \leq \mathcal{H}$ a subspace of a Hilbert space (or generally any normed space) is open, then it contains a ball around the origin $0 \in B_r(0) \subset M$, but for every (none-zero) vector $v \in \mathcal{H}$, we have $$ \frac{r}{2\Vert v \Vert} v \in B_r(0) \subset M $$ But M is a linear subspace so $ v \in M $. Thus the only open subspaces of $ \mathcal{H} $ are $ \mathcal{H} $ itself.

No.

Let $N$ be a normed space and

$M \subsetneq N \tag 1$

a proper subspace. Then $M$ contains no nonempty open set. For if

$\emptyset \ne U \subset M \tag 2$

were open, with

$M \ni m \in U, \tag 3$

we could find $\rho > 0$ such that the open ball

$B(m, \rho) \subset U; \tag 4$

then picking any

$0 \ne v \in N \setminus M \tag 5$

the vector

$m + \alpha (v - m) \in B(m, \rho) \tag 6$

if $0 \ne \alpha \in \Bbb R$ is sufficiently small, since

$\Vert (m + \alpha (v - m)) - m \Vert = \Vert \alpha (v - m) \Vert = \vert \alpha \vert \Vert v - m \Vert < \rho \tag 7$

for

$\vert \alpha \vert < \dfrac{\rho}{\Vert v - m \Vert}; \tag 8$

but then

$m + \alpha(v - m) \in M, \tag 9$

whence

$\alpha(v - m) = m + \alpha(v - m) - m \in M, \tag{10}$

whence

$v - m \in M, \tag{11}$

whence

$v = v - m + m \in M, \tag{12}$

in contradiction to (5); therefore no $B(m, \rho)$ as in (5) can exist, and $M$ cannot be open, since it contains no open set.