Let me first define some notations for convenience. Let $H$ be a Hilbert space and $\mathscr{I_\infty}(H)$ denotes the space of all compact operators on $H$. For $T\in\mathscr{I_\infty}(H)$ let $s_1(T)\ge s_2(T)\ge\cdots$ be a complete enumeration, multiplicity, of positive eigenvalues of $|T|:=\sqrt{T^*T}$. Then we have $$\displaystyle{s_n(T)=\min\limits_{\text{dim}(S)=n-1}\max\limits_{u\in S^\perp,\lVert u\rVert=1}\lVert Tu\rVert}$$
An operator $T\in\mathscr{I_\infty}(H)$ is said to be a trace class operator if $\lVert T\rVert_1:=\sum s_j(T)<\infty$ and we denote the space of all trace class operators on $H$ by $\mathscr{I_1}(H)$. We define, for $T\in\mathscr{I_1}(H)$, $\text{tr}\ T:=\sum\limits_j\langle e_j,Te_j\rangle$. I have proved that $$sup\{|\text{tr}\ TX|: X\in\mathscr{B}(H),\lVert X\rVert=1\}=\lVert T\rVert_1$$
The proof of $\mathscr{I_1}(H)$ forms a Banach space w.r.to. $\lVert\cdot\rVert_1$ goes as follows (from the book An Introduction to Quantum Stochastic Calculus by K.R. Parthasarathy):
I understood that $T$ is trace class and the sequence of bounded linear functionals $X\mapsto\text{tr}\ T_nX$ on $\mathscr{B}(H)$ is a cauchy sequence, hence it converges to some $\lambda\in\mathscr{B}(H)^*$ i.e. $$\text{sup}\{|\text{tr}\ T_nX-\lambda(X)|:\ X\in\mathscr{B}(H),\lVert X\rVert=1\}\to0$$
Now choosing $X=|u\rangle\langle v|$ with $\lVert u\rVert=\lVert v\rVert=1$, we have $\lambda(X)=\text{tr}\ TX$. This shows that $\lambda(X)=\text{tr}\ TX$ for finite rank operator $X$, hence by continuity of $\lambda$ (with $\lVert\cdot\rVert$ on $\mathscr{B}(H)$), it follows that $\lambda(X)=\text{tr}\ TX$ for all $X\in\mathscr{I_\infty}(H)$. For the completion of the proof, this much is enough because I can prove that $$\text{sup}\{|\text{tr}\ TX|:\ X\in\mathscr{I_{\infty}}(H),\lVert X\rVert=1\}=\lVert T\rVert_1$$
Applying this, $$\lVert T_n-T\rVert_1=\text{sup}\{|\text{tr}\ (T_n-T)X|:\ X\in\mathscr{I_\infty}(H),\lVert X\rVert=1\}=\text{sup}\{|\text{tr}\ T_nX-\lambda(X)|:\ X\in\mathscr{I_\infty}(H),\lVert X\rVert=1\}\to0$$
This finishes our proof, but my question is why $\lambda(X)=\text{tr}\ TX$ for all $X\in\mathscr{B}(H)$? I know that finite rank operators are dense in $\mathscr{B}(H)$ with respect to strong operator topology. Is this $\lambda$ here continuous with respect to strong operator topology?
Can anyone help me in this regard? Thanks for your help in advance.
Like you said, you don't need this to complete the proof, so ignore it. It's enough to conclude that $\lambda(X)=\operatorname{tr}TX$ for all $X\in \mathscr{I}_\infty(H)$ and then use $$\sup\{|\operatorname{tr}SX|:X\in\mathscr{I}_\infty(H),\|X\|=1\}=\|S\|_1$$ for all $S\in \mathscr{I}_1(H)$.
The reason that it's true is that from this and possibly further material $$\mathscr{I}_\infty(H)^* \simeq \mathscr{I}_1(H)$$ $$\mathscr{I}_1(H)^* \simeq \mathscr{B}(H)$$ via the pairing $(T,X)\mapsto\operatorname{tr}(TX)$. So $$\mathscr{I}_1(H)^{**} \simeq \mathscr{B}(H)^*$$ Since $T_n\rightarrow T$ in the norm of $\mathscr{I}_1(H)$, and a Banach space is canonically identified with a closed subspace of its double-dual, it follows that $T_n\rightarrow T$ when they are viewed as functionals on $\mathscr{B}(H)$ by the action $X\mapsto\operatorname{tr}(TX)$.