Suppose $H$ is any separable infinite-dimensional Hilbert space. Then $H$ has family of closed subspaces $\big\{ E_t :~ t \in [0,1]\big\}$ such that $E_s$ is a strict subspace of $E_t$ for all $0 \leq s < t \leq 1$.
Is my proof OK ?
Since $H$ is infinite-dimensional and separable, $H$ has a countable orthonormal basis (I know how to prove this). Consider $2^{\aleph_0}$ the space of binary sequences endowed with the following order : $$ f ~\prec~ g~~~~~~~~~~\Longleftrightarrow~~~~~~~~~~ f(n_{f,g}) ~<~ g(n_{f,g}),$$ where $n_{f,g}$ the least positive integer such that the $n$-th value of $f$ and $g$ differ.
For $f \in 2^{\aleph_0}$ define $E_f := \overline{\text{Span}\Big\{ e_n :~~ f(n) = 1 \Big\}}$. Then $f \prec g$ implies that $E_f$ is a closed proper subspace of $E_g$.
As Mathematician42 pointed out, the spaces you constructed are not nested.
A typical approach to this problem is to use the fact that $H$ is isomorphic to $L^2([0,1])$. Then $E_s=\{f\in L^2 : f(x)=0 \ \forall x>s\}$ works.
If you don't want to move to the space of functions, put the basis elements in bijection with the vertices of an infinite binary tree. There are uncountably many infinite branches of the tree, which are naturally ordered. You can let $E_s$ be the span of all elements to the left of the branch leading to binary number $s$.