Question on how to construct a confidence interval Consider a $K\times 1$ vector of  parameters, $\gamma\in \Gamma\subseteq \mathbb{R}^K$ . 
Consider the sequences $\{a_n\}_{n\in \mathbb{N}}$ and $\{T_n\}_{n\in \mathbb{N}}$, where $a_n$ is a $K\times 1$ vector of real numbers, $T_n\equiv \frac{1}{n}\sum_{i=1}^n X_i$, and $X_i$ is a random variable with support $\mathcal{X}$ for each $i=1,...,n$. 
Assume that:
1) $\{X_i\}_{i=1}^n$ are i.i.d. and $E(X_i)=0$, for each $n\in \mathbb{N}$. 
2) For each $n\in \mathbb{N}$, there is a realisation $t_n$ of $T_n$ such that 
$$
a_n'\gamma+t_n \geq 0
$$
3) We don't know the behaviour of the sequence $\{a_n\}_{n\in \mathbb{N}}$ as $n\rightarrow \infty$. However, we know that such sequence is bounded. 
Fix $n\in \mathbb{N}$ "large" and consider the set
$$
\Gamma^0_n\equiv \{\gamma\in \Gamma: a_n'\gamma+t_n \geq 0\}
$$
Suppose that we know the value of $a_n$ but we do not know the value of $t_n$. Then, by assumption 1) and for $n$ large,
$$
\tilde{\Gamma}^0_n\equiv \{\gamma\in \Gamma: a_n'\gamma \geq 0\}
$$
is a good estimate (not sure how to formalise this in terms of consistency) of $\Gamma^0_n$. 

Question: how we could think of getting a confidence region for a $\gamma\in \Gamma^0_n$? This confidence region should somehow take into account the fact that for $n$ not large enough, $t_n$ may be still far from $0$.
 A: A reasonable approach for this kind of problem is to look at the limits of the set difference between the estimator and the true set shrinks to the empty set, in some appropriate probabilistic sense, as $n \rightarrow \infty$.  To do this, we define the set $\Delta_n \equiv \tilde{\Gamma}_n - \Gamma_n = \{ \boldsymbol{\gamma} \in \Gamma | 0 \leqslant \mathbf{a}_n \cdot \boldsymbol{\gamma} < -T_n \}$ to be the difference between the estimated set and the actual set of interest. 
 (For simplicity, I have omitted the superscript from these sets, and I have changed notation to indicate vectors using bold letters.  Note also that this set difference depends on the random variable $T_n$ and so it is a random set.)
It is possible to use assumptions similar to those you have stated to obtain probabilities for the limit infimum and limit supremum of the above set difference.  We will show the first of these only.  To facilitate our analysis, let $\Gamma_i^* \equiv \{ \boldsymbol{\gamma} \in \Gamma | \mathbf{a}_i \cdot \boldsymbol{\gamma} \geqslant 0 \}$ so that $\Delta_i = \Gamma_i^* \cap \{ \boldsymbol{\gamma} \in \Gamma | T_i \leqslant - \mathbf{a}_i \cdot \boldsymbol{\gamma} \}$.  Then the limit infimum of the set difference can be written as:
$$\begin{equation} \begin{aligned}
\underset{n \rightarrow \infty}{\text{lim inf}} \ \Delta_n 
&= \bigcup_{n=1}^\infty \bigcap_{i=n}^\infty \Delta_i \\[6pt]
&= \bigcup_{n=1}^\infty \Bigg( \bigcap_{i=n}^\infty \Gamma_i^* \cap \{ \boldsymbol{\gamma} \in \Gamma | T_i \leqslant - \mathbf{a}_i \cdot \boldsymbol{\gamma} \} \Bigg) \\[6pt]
&= \Bigg( \bigcup_{n=1}^\infty \bigcap_{i=n}^\infty\Gamma_i^* \Bigg) \cap \Bigg( \bigcup_{n=1}^\infty \bigcap_{i=n}^\infty \{ \boldsymbol{\gamma} \in \Gamma | T_i \leqslant - \mathbf{a}_i \cdot \boldsymbol{\gamma} \} \Bigg) \\[6pt]
\end{aligned} \end{equation}$$
Since the sequence $\{ \mathbf{a}_i \}$ is bounded, there is some $0<a< \infty$ such that $\sup_i ||\mathbf{a}_i|| \leqslant a$.  It follows that: 
$$\begin{equation} \begin{aligned}
\bigcap_{i=n}^\infty \{ \boldsymbol{\gamma} \in \Gamma | T_i \leqslant - \mathbf{a}_i \cdot \boldsymbol{\gamma} \}
&\subseteq \bigcap_{i=n}^\infty \{ \boldsymbol{\gamma} \in \Gamma | T_i \leqslant - a \cdot ||\boldsymbol{\gamma}|| \}. \\[6pt]
\end{aligned} \end{equation}$$
Your condition (2) is a deterministic condition on the sequence of values $t_n$, and it is a very strong condition.  I will use the weaker assumption that $T_n$ has support over the whole real line, so $F_T(t) < 1$ for all $t \in \mathbb{R}$.  Under this condition, and using the fact that the random variables $T_1,T_2,T_3, ...$ are independent, we have:
$$\begin{equation} \begin{aligned}
\mathbb{P} \Bigg( \bigcap_{i=n}^\infty \{ \boldsymbol{\gamma} \in \Gamma | T_i \leqslant - \mathbf{a}_i \cdot \boldsymbol{\gamma} \} \Bigg)
&= \prod_{i=n}^\infty \mathbb{P} ( T_i \leqslant - \mathbf{a}_i \cdot \boldsymbol{\gamma} ) \\[6pt]
&\leqslant \prod_{i=n}^\infty \mathbb{P} ( T_i \leqslant - a \cdot ||\boldsymbol{\gamma}|| ) \\[6pt]
&= F_T (- a \cdot ||\boldsymbol{\gamma}|| )^\infty = 0. \\[6pt]
\end{aligned} \end{equation}$$
Since each of these events (for all $n$) have probability zero, it follows that:
$$\begin{equation} \begin{aligned}
\quad \quad \quad \ \ \
\mathbb{P} \Bigg( \bigcup_{n=1}^\infty \bigcap_{i=n}^\infty  \{ \boldsymbol{\gamma} \in \Gamma | T_i \leqslant - \mathbf{a}_i \cdot \boldsymbol{\gamma} \} \Bigg)
&\leqslant \sum_{n=1}^\infty \mathbb{P} \Bigg( \bigcap_{i=n}^\infty  \{ \boldsymbol{\gamma} \in \Gamma | T_i \leqslant - \mathbf{a}_i \cdot \boldsymbol{\gamma} \} \Bigg) \\[6pt]
&= \sum_{n=1}^\infty 0 = 0. \\[6pt]
\end{aligned} \end{equation}$$
This establishes that $\mathbb{P}({\text{lim inf}}_{n \rightarrow \infty} \ \Delta_n) = 0$, which establishes a kind of weak convergence in set-theoretic terms.  If you were to conduct an analogous derivation on the limit supremum, you would find that this has probability one under these same assumptions, so you have a kind of indeterminate situation.  Note that in this working I would used a different (weaker) assumption than your equation (2), and I have not required the assumption that the expectation of $X_i$ is zero.
