Suppose $D_1$ is a unit disk in $\mathbb{R}^2$ and $u_n \in H^1(D_1;\mathbb{R})$. We assume further that $\|u_n\|_{H^1} <1$ and $u_n = 0$ on $A:=\big\{(x_1,x_2) : x_1 \in [-1,1] \hspace{2pt} \text{ and } \hspace{2pt} x_2=0 \big\}$ $\mathcal{H}^1$-almost everywhere (this is possible by trace theorem). Here $\mathcal{H}^1$ is the one-dimensional Hausdroff measure.
It is clear that there is a limiting map $u_\infty \in H^1(D_1;\mathbb{R})$ such that $u_n$ converges to $u_\infty$ weakly in $H^1(D_1)$, up to a subsequence. By Fatou's Lemma, $$ \int_0^1 \liminf_{n} \int_0^{2\pi} |\nabla u_n|^2 r d\theta dr \leq \liminf_{n} \int_0^1 \int_0^{2\pi} |\nabla u_n|^2 r d\theta dr <1.$$
Therefore, we can find a positive number $R \in (1/2,1)$ such that
$$ \begin{split} \int_0^{2\pi} |\partial_\theta u_\infty(R,\theta)|^2 R^{-1} d\theta & \leq\liminf_{n} \int_0^{2\pi} |\partial_\theta u_n(R,\theta)|^2 R^{-1} d\theta \\ & =\liminf_{n} \int_0^{2\pi} |\nabla u_n|^2 \Big|_{r=R} R d\theta <1. \end{split} $$ We can obtain that $$ \int_0^{2\pi} |\partial_\theta u_n(R,\theta)|^2 d\theta < 4 \hspace{10pt} \text{for all $n$ large enough.}\label{1}\tag{1} $$
My question is, can we also find a positive number $\sigma \in (1/2,1)$ such the above inequality \eqref{1} holds and $u_n=0$ at the point $(\sigma,0)$ for all $n$ large?
I am trying to use the equi-continuity of $u_n$. Using \eqref{1} and the fundamental theorem of calculus, we see that $u_n$ is equi-continuous on $\partial D_{R}$. Since
$$ | u_n(R,\theta_1) - u_n(R,\theta_2)| \leq |\theta_1 - \theta_2|^{1/2} \left(\int_0^{2\pi} |\partial_\theta u_n(R,\theta)|^2 d\theta\right)^{1/2} < 4 |\theta_1 - \theta_2|^{1/2}.$$ But I do know how to get the desired result.
In order to avoid technical difficulties, let me straighten out the picture. For your specific case, I included a remark at the end.
$$ u_n \in H^1((-1,1)^2), \quad \| u_n \|_{H^1} \le 1, \quad u_n = 0 \text{ on } A = \{ y=0 \} \text{ in the sense of trace.} $$ And say we're looking for a slice $I_x = \{ x \} \times (-1,1)$ (given by some $x \in (-1,1)$) such that $u|_{I_x}$ is in $H^1(I_x)$, $\| u|_{I_x} \|_{H^1} \le 1$ and $u(x,0) = 0$ in the sense of trace.
You already observed that $u|_{I_x}$ is in $H^1(I_x)$ for almost every $x \in (-1,1)$. Moreover, the weak derivative $u|_{I_x}$ is given by the restriction of $\partial_y u$ to $I_x$ (you haven't mentioned this, but it's true). You also showed (up to some technicalities) that $\| u|_{I_x} \|_{H^1} \le 1$ on some positive set of $x \in (-1,1)$. The problem is in justifying $u(x,0) = 0$.
$\newcommand{\pl}{\partial}$ $\newcommand{\R}{\mathbb{R}}$ $\newcommand{\tr}{\operatorname{tr}}$
The solution could be to define the trace in this way. That is, for good $x \in (-1,1)$ as above, we know that $u(x,0)$ is well-defined in the sense of trace. One can even choose a continuous representative of $u_{I_x}$. The important part is that $$ |u(x,0)| \le C \| u|_{I_x} \|_{H^1}, $$ with some uniform constant $C$. This can be done for almost all $x \in (-1,1)$, so we have thus defined a function $\tr u \colon (-1,1) \to \R$ (defined a.e.). Let us check that $\tr u \in L^2((-1,1))$: $$ \int_{-1}^1 |\tr u(x)|^2dx = \int_{-1}^1 |u(x,0)|^2 dx \le C^2 \int_{-1}^1 \| u|_{I_x} \|^2_{H^1} dx = C^2 \| u \|_{H^1((-1,1)^2)}, $$ where the last equality is a byproduct of our proof that $u_{I_x} \in H^1$ for a.e. $x$.
Let us summarize:
Due to density of $C^1_c$ in $H^1$, there is exactly one such operator, and it is the trace operator.
In our case, if $u \in H^1((-1,1)^2)$, we can be sure that our two ways of looking at $u(x,0)$ - trace of $u$ evaluated at $x$ versus trace of $u_{I_x}$ - coincide. If one is zero, the other one is too (say, for a.e. $x$).
In your specific case of polar coordinates, I see at least two options. First, one could just repeat the same calculations with $rdrd\theta$ instead of $dxdy$. Second, one could use a diffeomorphism that straighens out our circles. In both cases, one has to be careful about the point $(0,0)$, so it might be useful to restrict our attention to an annulus $\varepsilon < r < 1$ (at least for a moment).