On the isometry between bounded linear operators and the dual of nuclear linear operators

I would do it as follows:

1. For $(x,y) \in H \times H$ define an element of $N(H)$ by $x \,\tilde{\otimes}\, y = \langle \cdot , y \rangle x$ and note that $\|x \, \tilde{\otimes}\,y\|_{1} = \|x\|\,\|y\|$.
2. For a functional $\varphi \in N(H)'$ define a sesquilinear form on $H$ by $B_{\varphi}(x,y) = \varphi(x \,\tilde{\otimes}\, y)$. The map $\varphi \mapsto B_{\varphi}$ is clearly linear and of norm $\leq 1$ since $|B_{\varphi}(x,y)| = |\varphi(x \, \tilde{\otimes}\,y)| \leq \|\varphi\|\,\|x\|\,\|y\|$ by 1.
3. Use the version of the Riesz representation theorem stating that every bounded sesquilinear form $B$ is of the form $B(x,y) = \langle x,T y \rangle$ for a unique $T$. Moreover, $\|B\| = \|T\|$. In other words, the map $B \mapsto T$ from bounded sesquilinear forms to $L(H)$ is a linear isometry.
4. Define $j(\varphi)$ by $\varphi \mapsto B_{\varphi} \mapsto T_{\varphi}$ by combining the maps from 2. and 3. Verify that $ji = 1_{L(H)}$ and $ij = 1_{N(H)'}$. Since both $i$ and $j$ have norm $\leq 1$, they must be isometries.