For any bounded operator T on H there exists a sequence of finite rank operators $(T_n)_{n=1}^{\infty}$ such that $T_n → T$ strongly. I have come up with a proof which I want to confirm as some parts of it make me unsure whether its correct or not.
Note: $T_n \to T$ strongly means 
$$\|T_n(x)-T(x)\| \to 0 \ \ \ \forall x\in H.$$
Here is my proof attempt.
Let $(e_j)_{j=1}^{\infty}$ be an orthonormal basis for $H$ ($H$ is assumed separable) and $(g_k)_{k=1}^{\infty}$ be a sequence consisting of finite linear combinations of $e_j$, more precisely for each g in H, we choose 
$$g_k= \sum_{i=1}^{k} \langle g,e_i\rangle  e_i.$$ 
Now define $T_n$ by 
$$T_n(g_k)=\begin{cases} T(g_k) & \text{for }k\leq n, \\ 0 &\text{for } k>n.\end{cases}$$
Clearly each $T_n$ has finite rank since the range of each $T_n$ is the span of $(g_1,..,g_n)$. Also $||T_n (g_k)-T(g_k)||=0$ for $n \geq k$ Let now g be an arbitrary element of H. Then since $(g_k)_{k=1}^{\infty}$ is dense in H, $||g-g_k||\to 0$ as $k\to \infty$. Then by the triangle inequality, 
$$||T_n(g)-T(g)||\leq ||T_n(g)-T_n(g_k)|| + ||T_n(g_k)-T(g)||$$
First taking the limit as $n\to \infty$, we get that the first term goes to 0 (by continuity of norm) since $T_n(g_k)=T_n(g)$ for $n\geq k$, and the second term goes to $||T(g_k)-T(g)||$. Now 
$$||T(g_k)-T(g)||\leq \|T\| \|g_k-g\|$$ 
by definition of operator norm. Since $||T||<\infty$, taking the limit as k goes to $\infty$ and using the fact that $||g_k-g||\to 0$ we get that $||T_n(g)-T(g)||\to 0$ as $n\to\infty$
My question is, the definition my book uses for a basis is that $(e_n)$ is a basis if finite linear combinations are dense (in the norm) in H. Now for me that means for each g in H,given $\epsilon>0$ there exists a $(g_k)$ which is a finite linear combination of $e_j$ such that $||g_k-g||< \epsilon$. Is that equivalent to the existence of a sequence $(g_k)$ where each $g_k$ is a finite linear combination of $e_j$'s such that $||g_k-g||\to 0$ as $k\to \infty$ ? Also, is my proof correct? Taking limits with respect to k and n bothers me a bit whether I am allowed to do that or not. Thanks in advance for any comments.
Edit: Reading my notes, it seems I need to be more precise on what the $g_k$ are. They depend on the g that would be chosen. More precisely, $g_k= \sum_{i=1}^{k} <g,e_i> e_i$. Then it works?
 A: (Migrated from comments)
Let $P_n g = \sum_{k=1}^n \langle e_k, g \rangle e_k$ so that your $g_n$ is written by $g_n = P_n g$. It is easy to check that each $P_n$ is a bounded linear operator of finite rank. (Here I am assuming that $\langle \cdot, \cdot \rangle$ is linear in the second argument and conjugate-linear in the first argument.)
Then we define $T_n$ by
$$ T_n g = TP_n g. $$
Since $P_n$ has finite rank, the same is true for $T_n$. Moreover,
$$ \| T_n g - T g \| = \| T(P_n g - g) \| \leq \|T\| \|P_n g - g\| $$
Since $P_n g \to g$ as $n \to \infty$, it follows that $T_n g \to T g$.

Addendum. If you are also required to establish the convergence $P_n g \to g$, here is a proof:

Lemma. Let $S = \{e_1, e_2, \cdots, e_n\}$ be an orthonormal subset of $H$ and define $P g$ by
  $$P g = \sum_{k=1}^n \langle e_k, g \rangle e_k.$$
  Then for any $h \in \operatorname{span}(S)$, we have
  $$ \| g - h \|^2 = \| g - P g\|^2 + \|P g - h \|^2. $$

Indeed, using the fact that $P g - h \in \operatorname{span}(S)$ we can write $P g - h = \sum_{k=1}^{n} c_k e_k$ for some constants $c_1, \cdots, c_n$. Then
$$ \langle g - P g, P g - h \rangle
= \sum_{k=1}^{n} c_k \langle g - P g, e_k \rangle
= \sum_{k=1}^{n} c_k (\langle g, e_k \rangle - \langle g, e_k \rangle )
= 0. $$
Then $\text{(*)}$ follows by plugging this observation to the identity
$$ \| g - h \|^2
= \| g - P g \|^2 + \underbrace{2\operatorname{Re}\langle g - P g, P g - h \rangle}_{=0} + \| P g - h \|^2. $$
Now we return to the proof. For each $\epsilon > 0$, pick $h \in \operatorname{span}\{e_1, e_2, \cdots \}$ so that $\| g - h \| < \epsilon$ holds. Choose $N$ so that $h \in \operatorname{span}\{e_1, \cdots, e_N\}$. Then for any $n \geq N$, we also have $h \in \operatorname{span}\{e_1, \cdots, e_n\}$ and hence
$$ \| g - P_n \| \leq \| g - h \| < \epsilon $$
by Lemma. This proves that $P_n g \to g$ as $n \to \infty$.
