First of all, in order to have $(iii)$ be equivalent to $(i)$ and $(ii)$, we need that $Y$ is complete (a Banach space) too.
As a counterexample when $Y$ is not complete, consider $X = \ell^p$, and $Y$ the subspace of sequences with only finitely many nonzero terms. Then let
$$(T_nx)_k = \begin{cases}x_k &, k \leqslant n\\ 0 &, k > n. \end{cases}$$
$T_n$ is a bounded sequence of linear operators $X\to Y$, and $T_n(x)$ converges for all $x$ in the dense subspace $Y$ of $X$. But $T_n(x)$ does converge only for $x\in Y$, so neither $(i)$ nor $(ii)$ hold in this example.
Your attempt to prove $(i) \Rightarrow (iii)$ does not work, there are two problems. First, from $\lVert T_n(x)\rVert < N$, you cannot deduce $\lVert T_n\rVert_{op}\lVert x\rVert < N$, since you only have the inequality $\lVert T_n(x)\rVert \leqslant \lVert T_n\rVert_{op}\lVert x\rVert$, not equality.
Second, the bound on the sequence $T_n(x)$ depends on $x$, you only have $\lVert T_n(x)\rVert \leqslant N(x)$.
To show the uniform boundedness, you consider the closed sets $$A_K = \{x \in X : \lVert T_n(x)\rVert \leqslant K \text{ for all } n\}.$$
Since $T_n(x)$ is bounded for every $x\in X$, you have
$$X = \bigcup_{k=1}^\infty A_k.$$
From Baire's theorem, deduce that some $A_k$ is a neighbourhood of $0$ in $X$.
For the implication $(iii) \Rightarrow (ii)$ (under the assumption that $Y$ is complete!), note that
- the set $C = \left\{ x \in X : \lim\limits_{n\to\infty} T_n(x)\text{ exists}\right\}$ is a linear subspace of $X$,
- the map $T\colon C \to Y$ defined by $T(x) = \lim\limits_{n\to\infty} T_n(x)$ is linear and continuous.
Then the general theory asserts the existence of a continuous (linear) extension $\tilde{T}\colon \overline{C}\to Y$. Since $C$ is by assumption dense, $\overline{C} = X$. Now use the boundedness of the norms - the equicontinuity of the family $(T_n)$ - to deduce that actually $C = X$.
Best Answer
Take the limit of the inequality $\|T_n x_j\| \le k$ when $j\to +\infty$.