Let $X$, $Y$ be Banach spaces and $\{T_{j,k} : j,k \in\Bbb N\}$ be bounded linear maps from $X$ to $Y$. Suppose that for each $k$ there exists $x\in X$ such that $\sup\{\lVert T_{j,k} x\rVert : j \in\Bbb N\} =+\infty$. Then there is an $x$ such that $\sup\{\lVert T_{j,k} x\rVert : j \in\Bbb N\} =+ \infty$ for all $k$.
How can i argue by contradiction, and apply Baire categories theorem to
$$ F_{k,n}=\bigcap_{j\in \mathbb{N}} \{x\in X , \lVert T_{j,k} (x)\rVert_Y \leq n\} $$
So \begin{align*} X=\bigcup_{k,n}F_{k,n}, \end{align*} and each $F_{k,n}$ is closed, so by Baire Theorem some $k,n$ are such that $F_{k,n}^{\circ}\ne\emptyset$, so some $x\in X$, $\delta>0$ are such that \begin{align*} B_{\delta}(x)\subseteq F_{k,n}. \end{align*} For this $k$, we have by assumption some $x_{k}$ is such that $\sup\{\|T_{j,k}(x_{k})\|:j\in\mathbb{N}\}=\infty$.
However, we have \begin{align*} x-\dfrac{x_{k}}{\delta(\|x_{k}\|+1)}\in B_{\delta}(x), \end{align*} and that \begin{align*} x-\dfrac{x_{k}}{\delta(\|x_{k}\|+1)}+\dfrac{x_{k}}{\delta(\|x_{k}\|+1)}=x. \end{align*} We also have \begin{align*} \left\|T_{j,k}\left(x-\dfrac{x_{k}}{\delta(\|x_{k}\|+1)}\right)\right\|\leq n, \end{align*} and that \begin{align*} \|T_{j,k}(x)\|\leq n, \end{align*} so by triangle inequality, it follows that \begin{align*} \left\|T_{j,k}\left(\dfrac{x_{k}}{\delta(\|x_{k}\|+1)}\right)\right\|\leq 2n, \end{align*} so \begin{align*} \|T_{j,k}(x_{k})\|\leq 2n\delta(\|x_{k}\|+1), \end{align*} varying $j\in\mathbb{N}$ arbitrarily, then \begin{align*} \sup\{\|T_{j,k}(x_{k})\|: j\in\mathbb{N}\}\leq 2n\delta(\|x_{k}\|+1)<\infty, \end{align*} a contradiction.