Problem: Suppose that you have a separable Hilbert space, $H$, an orthonormal basis of $H$, $(e_n)$, and the set $D=\{T\in B(H):T\,\text{is diagonal with respect to the basis}\,(e_n)\}$. Show that $(D,\|\cdot\|_{op})$ is a unital Banach algebra.
My Attempt: I am mainly having issues with establishing that $(D,\|\cdot\|_{op})$ is complete, and would appreciate some pointers to get me going in the right direction.
I recall the definition I am using for a unital Banach algebra:
"A unital Banach algebra" $(A,\|\cdot\|)$ is an algebra $A$ over $\mathbb C$ with identity $1$ and a norm $\|\cdot\|$ such that $(A,\|\cdot\|)$ is a Banach space, $\|ab\|\le\|a\|\|b\|$ and $\|1\|=1$"
I begin by establishing that the operation of multiplication on $D$ is submultiplicative and that the identity element of $D$ has unit norm.
For $x\in H$ and $S,T\in D$, consider: $\|(S\circ T)x\|=\|S(Tx)\|\le\|S\|\|Tx\|\le\|S\|\|T\|\|x\|$. And this allows us to deduce that $\|(S\circ T)\|\le\|S\|\|T\|$, thus establishing the submultiplicativity of multiplication on $D$. This is well defined as can be seen here: What does it mean for an operator to be diagonal with respect to an orthonormal basis? - my thanks to those who helped me there!
The identity element $1$ on $D$ is, as it is on $B(H)$, given by identity operator on $H$, where, for $x\in H$, we have that $I_Hx=x$. Consider then, $\|I_H\|=\sup_{\|x\|=1}\|I_Hx\|=\sup_{\|x\|=1}\|x\|=1$, which is what we needed to show.
All that is left to do is to show that $(D,\|\cdot\|_{op})$ is complete.
Take a Cauchy sequence, $(T_n)\in D$ and assume that $T_n\to T\in B(H)$. The strategy is to show that this $T\in B(H)$ is actually in $D$. Consider, $\|T_n-T_m\|$ and let $m\to\infty$ so as to consider $\|T_n-T\|$. Now,
$$\|T_n-T\|=\sup_{\|x\|=1}\|(T_n-T)x\|=\sup_{\|x\|=1}\|T_nx-Tx\|$$
Now I know that if $H$ is separable then $H$ has a countable orthonormal basis. I also know that given an orthonormal basis for $H$ then $x=\sum_{i=1}^{\infty}(x,e_i)e_i$ for all $x\in H$. I think it is these facts I am trying to make use of here. Now,
$$T_nx=T_n\sum_{i=1}^{\infty}(x,e_i)e_i=\sum_{i=1}^{\infty}(x,e_i)T_ne_i$$
And since $T_n\in D$, for all $n\in \mathbb N$, we have that $\sum_{i=1}^{\infty}(x,e_i)T_ne_i=\sum_{i=1}^{\infty}(x,e_i)\lambda_ie_i$.
But where do I go from here? The continuity of $T$ will allow me to take $T$ inside of the series but nothing else as far as I can see. I also haven't made use of the fact that $H$ being separable ensures that $(e_n)$ is countable. Is this the wrong way to use that $H$ is separable?
A direct way to address your problem is, as Matthias said, note that norm convergence implies implies pointwise convergence.
Here is a different approach. One can show that $D'=D$, that is the commutant of $D$ is $D$ itself. Indeed, if $ST=TS$ for all $T\in D$, in particular $SE_{k}=E_{k}S$ for all $k$, where $E_k=\langle\cdot,e_k\rangle e_k$ is the orthogonal projection onto the span of $e_k$. Then $$ Se_k=SE_ke_k=E_kSe_k=\lambda_k\,e_k. $$ So $S\in D$. This shows that $D'\subset D$. Since $D$ is abelian, we also have $D\subset D'$.
Thus $D=D'$. It is not hard to check that the commutant of a $*$-algebra is a von Neumann algebra: a C$^*$-algebra that is also closed in the wot and sot topologies. In particular, it is also norm closed.