This is the corollary under consideration, which I have proved:
If $\left\{A_n\right\}_{n\in\mathbb N}$ is a sequence of continuous, linear maps from an $F$-space $X$ to a topological vector space $Y$, and $\lim_{n\to\infty}A_nx=Ax$ exists for all $x\in X$, then $A$ is linear and bounded.
How is this not a counterexample?
Define $A_n:c_0\to c_0$ by
$$\left(A_nx\right)_m=\begin{cases} mx_m&m\leq n,\\ 0&m>n, \end{cases}$$
where $c_0$ is the space of all real sequences that converge to $0$. Then $x=\left(m^{-1/2}\right)_{m\in\mathbb N}\in c_0$ but $\left(Ax\right)_m=m^{1/2}\notin c_0$ since $\left\|Ax\right\|_\infty=\infty$.
Your example doesn't satisfy the hypothesis that $\lim_{n\to\infty}A_nx$ exists for all $x$. Indeed, for your choice of $x=(m^{-1/2})$, $A_nx$ does not converge to any element of $c_0$. (The sequences $A_nx$ converge pointwise to the sequence $(m^{1/2})$, but this sequence is not an element of $c_0$ and in any case it does not converge with respect to the sup norm.)