I will try to answer in a "how to keep things in mind way".
First of all, the most common and most flexible definition of separability is the first: A topological space $X$ is called separable when there exists a countable subset $\{x_n\}_{n=1}^\infty$ that is dense in the space. Now we don't care about topology in general, we care about Hilbert spaces, but the definition is the same.
Using Zorn's lemma, one proves that every Hilbert space admits an orthonormal basis. An orthonormal basis of a Hilbert space $H$ is a set of vectors $(e_i)_{i\in I}$ such that $\|e_i\|=1$ for all $i\in I$, $\langle e_i,e_j\rangle=0$ when $i,j\in I$ and $i\neq j$ and it is true that for each $x\in H$ it is
$$x=\sum_{i\in I}\langle x,e_i\rangle e_i$$
in the sense that for each $\varepsilon>0$ we may find a finite set $F_0\subset I$ such that for any finite set $F\subset I$ with $F_0\subset F$ it is
$$\bigg{\|}x-\sum_{i\in F}\langle x,e_i\rangle e_i\bigg{\|}<\varepsilon.$$
Don't let this scare you! Once you learn about nets this will be very easy to digest. Just keep in mind that in the case that our index-set $I$ is countable, the sum above is simply a series (in the way that even your cat understands what a series is).
An equivalent definition is this: $(e_i)_{i\in I}$ is an orthonormal basis for $H$ if and only if $\|e_i\|=1$ for all $i$, $\langle e_i,e_j\rangle=0$ for all $i\neq j$ and the following implication is true for any $x\in H$:
$$\bigg(\text{ for all i}\in I: \langle x,e_i\rangle=0\bigg)\implies x=0 $$
Anyway. It can be proved that if $(e_i)_{i\in I}$ and $(f_j)_{j\in J}$ are both orthonormal bases for a Hilbert space $H$, then $I$ and $J$ have the same cardinality. We say that this common cardinality is the Hilbert dimension of the Hilbert space.
One can easily convince themselves that a Hilbert space has finite Hilbert dimension if and only if it is finite dimensional in the usual sense.
Now one can also prove the following: a Hilbert space is separable if and only if its Hilbert dimension is (finite or) countably infinite.
Hint on how to prove the last thing: Obviously finite dimensional spaces are separable. If a space is separable, start from a dense sequence, pass to a dense linearly independent subsequence and apply Gram-Schmidt. Conversely, if a space has countably infinite Hilbert dimension, then the $\mathbb{Q}$-span of a countable orthonormal basis is a dense, countable set.
A good reference for all this is Conway's first Chapter. The exercises are difficult though, be warned about that!
Best Answer
The statement can't be true: let $\mathcal{H}$ be a separable Hilbert space. Since it is separable, consider a countable orthonormal basis $\{\alpha_1,\alpha_2,\dots\}$ for $\mathcal{H}$. For each bounded sequence $x=(x_n)$ of real numbers, that is an element of $l_{\infty}$, consider the diagonal operator on $\mathcal{H}$, defined by $$X_{x}(\alpha_n)= x_n\alpha_n$$ for $\alpha_n$ an orthonormal basis vector.
Then $\|X_x\|_\text{op} = \|x\|_{l_\infty}$ (check!), and this induces an isometric inclusion of $l_{\infty}$ into the space of all linear bounded operators on $\mathcal{H}$, and $l_{\infty}$ is not separable!