The weak operator topology means that $(T_nx,y)\to (Tx,y)$ for all $x,y\in H$, where $(\cdot,\cdot)$ is our inner product. This is certainly the case if we could show, for all $x,y\in H$ and $\varepsilon>0$, that there exists $N\in\mathbb N$ such that
$$
|(Tx-T_nx,y)|<\varepsilon
$$
for all $n>N$. (If that doesn't make sense, you should verify it). Well, by Cauchy-Schwarz,
$$
|(T_nx-Tx,y)|\leq\|T_nx-Tx\|\cdot\|y\|.
$$
Since $\|y\|=M$ is finite, we can choose $N\in\mathbb N$ such that $\|T_nx-Tx\|<\varepsilon/M$ for $n>N$, which in turn shows that
$$
(T_Nx-Tx,y)\leq |(T_Nx-Tx,y)|<\varepsilon
$$
for all $n>N$. Because this works for all $\varepsilon>0$, we have $(T_nx,y)\to(Tx,y)$ as $n\to\infty$.
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
You are using that $f$ is uniform continuous when restricted to each open $W$ so that $W\subset\subset U$.
Then for all $\delta >0$, there is $\epsilon_0 >0$ so that $$|f(x) - f(y)| <\delta$$ whenever $x, y\in W$ and $|x-y| < \epsilon_0$. Now if $x\in V$. Let $\epsilon_0$ be small so that $|y-x|<\epsilon_0 \Rightarrow y\in W$ (it is in general not true that $y\in V$). Then for all $\epsilon <\epsilon_0$,
$$\begin{split} |f(x) - f^\epsilon(x)| &= \left| f(x) - \int_U f(y) \eta_\epsilon (y-x) dy\right| \\ &\le \int_U |f(x) - f(y)| \eta_\epsilon (y-x) dy \\ &\le \delta \end{split}$$
Note that in the second inequality we are using the uniform continuity of $f$ on $W$, not on $V$. As $x\in V$ is arbitrary, $f^\epsilon \to f$ uniformly on $V$.
Remark: The compactness of $V$ is used to guarantee the existence of such a $W$. Indeed, let $d>0$ be such that $d(v, \partial U) > d$ for all $v\in V$ (This $d$ can be found as $V$ is compact. Then let $W = V_{d/2} = \{ x\in U : d(x, V) < d/2\}$. Then $W$ is open and $\overline W$ is compact and is in $U$.