Let $\{X_n\}^{\infty} _{n=1}$ be a sequence S.T, for some $N \in \mathbb{N}, X_n = \alpha$ for $n ≥ N$. Show $\{X_n\}^{\infty} _{n=1}$ converges. I'm not too sure how to approach this problem. Given the things at my disposal I'm not sure what I'd be able to use.
Theorem1: If a sequence $\{A_n\}^{\infty} _{n=1}$  converges, then it has a unique limit.
Theorem2: If the sequence $\{A_n\}^{\infty} _{n=1}$ converges, then the set $\{A_n| n \in \mathbb{N}\}$ is bounded.
plus the definition of convergence to a limit and its equivalent using neighborhood method.
 A: Just use the definition. I will talk about real sequences here such that we can make it more visual (I want the graph to be $2$-dimensional). Recall, that convergence means that the sequence gets arbitrarily close to some real number $a$. This is formalized as follows:
No matter how small the distance $\epsilon > 0$ is, you have to be able to find some point on the $x$-axis (given by some natural number $N$), such that all $x_n$ (which are the function values visually) with $n \geq N$ (i.e. all elements for the sequence starting from that chosen point) lie in $(a - \epsilon, a + \epsilon)$ (that means the distance between these $x_n$ and the limit $a$ is smaller than $\epsilon$ and is usually written as $\lvert x_n - a \rvert < \epsilon$).
Now let us take a look at your problem:
Take some $\epsilon > 0$ and choose $N \in \mathbb{N}$, such that $x_n = \alpha$ for all $n \geq N$. Thus we have that the sequence is constant (with value $\alpha$ if we only look at natural numbers $n \geq N$. This means, we expect that it converges towards $\alpha$. Now you only need to write that down formally. As $n \geq N$, we get
$$\lvert x_n - \alpha \rvert = \lvert \alpha - \alpha \rvert = 0 < \epsilon,$$
which means that $(x_n)_{n \in \mathbb{N}}$ converges with limit $\alpha$.
