Let $T\in(0,\infty]$ and $\Omega\subseteq\mathbb R^d$ be bounded and open. For $k\in\mathbb N$, let $I_k:=(0,\min(k,T))$, $u_k\in C(\overline{I_k}\times\overline\Omega)$ and $$T_k:=\min\left(\inf\left\{t\in\overline{I_k}:\left\|u_k(t,\;\cdot\;)\right\|_\infty\ge k\right\},k\right).$$ Assume $$u_k=u_{k+1}\;\;\;\text{on }\overline{I_k}\times\overline\Omega\tag1$$ for all $k\in\mathbb N$ and $(T_k)_{k\in\mathbb N}$ is nondecreasing such that $$T_0:=\lim_{k\to\infty}T_k$$ and $$u(t,x):=\lim_{k\to\infty}1_{[0,\:T_k]}(t)u_k(t,x)\;\;\;\text{for }(t,x)\in[0,T_0)\times\overline\Omega$$ are well-defined.
Assuming $T\ne T_0$, is it possible that $$c:=\limsup_{t\to T_0-}\left\|u(t,\;\cdot\;)\right\|_\infty<\infty?\tag2$$
My attempt is to assuming $(2)$ will lead to a contradiction to the definition of $u$. If $(2)$ holds, then there is clearly a nondecreasing sequence $(t_k)_{k\in\mathbb N}\subseteq[0,T_0)$ with $t_k\xrightarrow{k\to\infty}T_0$ and $$\sup_{k\in\mathbb N}\left\|u(t_k,\;\cdot\;)\right\|_\infty\le c.\tag3.$$ I guess we can choose the $t_k$ such that $t_k\in(0,T_k)$, can we? Now we somehow need to utilize that $u_k=u$ on $[0,T_k]\times\overline\Omega$ and yield a contradiction to the definition of $T_k$ and the continuity of $u_k$ for large enough $k$ ...