Let $N$ be a von Neumann algebra and $\varphi$ is a normal state on $N$. $U(N)$ is denoted by the set of all unitary elements in $N$. Put $$C_{\delta}(x)=\bar{co}^w\{uxu^*|u\in U(N),\|u\varphi-\varphi u\|＜\delta\}.$$
Suppose $y$ is the element of minimal $\|·\|_{\varphi}$-norm in $C_{\delta}(x)$ and $y_n\in C_{\delta}(x)$. If $\|y_n\|_{\varphi}\to \|y\|_{\varphi}$, since $y$ is the element of minimal $\|·\|_{\varphi}$-norm in $C_{\delta}(x)$, this forces $\|y_n-y\|_{\varphi}\to 0$.
If $y$ is the element of minimal $\|·\|_{\varphi}$-norm in $C_{\delta}(x)$, we have $\|y\|_{\varphi}=\inf\{\|z\|_{\varphi}, z\in C_{\delta}(x)\}$, But how to deduce that $\|y_n-y\|_{\varphi}\to 0$ by virtue of $\|y_n\|_{\varphi}\to \|y\|_{\varphi}$ and $y_n\in C_{\delta}(x)$?
Notation:$\|x\|_{\varphi}=\varphi(x^*x)^{\frac{1}{2}}$
I'm guessing that you meant $\|u\varphi-\varphi u\|<\delta$, though that's not relevant to the answer. The only relevant parts are that $C_\delta(x)$ is closed and convex in the $\varphi$-norm.
The point is that $[x,y]=\varphi(y^*x)$ is an inner product. In particular it satisfies the parallelogram identity. And the computation one needs is the one that shows the uniqueness of the distance minimizer to a closed convex set in a Hilbert space.
Namely, you have \begin{align} \|y_n-y\|_\varphi^2 &=2\|y_n\|_\varphi^2+2\|y\|_\varphi^2-\|y_n+y\|_\varphi^2\\[0.3cm] &=2\|y_n\|_\varphi^2+2\|y\|_\varphi^2-4\Big\|\frac{y_n+y}2\Big\|_\varphi^2\\[0.3cm] &\leq2\|y_n\|_\varphi^2+2\|y\|_\varphi^2-4\|y\|_\varphi^2\\[0.3cm] &\xrightarrow[n,m\to\infty]{}0. \end{align}