Given $\lbrace{ a_n \rbrace} \rightarrow a$, ie., the sequence $a_n$ converges in the limit to $a$, we have to prove the following relationship
$$\inf_n a_n \le a \le \sup_n a_n$$
Proof attempt. Since $\lbrace{ a_n \rbrace} \rightarrow a$, so for all $\varepsilon > 0$, $\lvert a_n - a \rvert < \varepsilon$.
Therefore, $-\varepsilon < a_n - a < \varepsilon \Rightarrow a -\varepsilon < a_n < a + \varepsilon$.
Now we prove the left half of the inequality in question (the right half is similar). Suppose for contradiction that $a < \inf_n a_n = \alpha$; then because $\alpha$ is the infimum, for $\varepsilon > 0$ we have,
$$a < \alpha \le a_n < a + \varepsilon < \alpha + \varepsilon $$
So this is the part I can't get past; I can't think of a way the above relationship can be contradictory. One line of reasoning is that for arbitrary choices of $\varepsilon$, we can construct any number of cases, such as with $\frac{\varepsilon}{2}$, so that $\alpha \le a + \frac{\varepsilon}{2} \le a_n < a + \varepsilon < \ldots$ which means I can always have a larger $\alpha$.
This type of reasoning, however, seems a little too convoluted to be useful. If I am on the correct track, how should I formalise this argument? Alternatively, if I am incorrect, where am I going wrong?
A few general comments about your answer specifically.
As was mentioned in a comment above, your statement of the definition of the limit of a sequence is incorrect.
$\alpha$ is not the supremum. It is the infimum, by your definition. I think that's just an honest mistake on your part.
Now, you've shown that:
$$a < \alpha \leq a_n < a+\epsilon < \alpha+\epsilon$$
But where does that bring you, though? It doesn't seem to establish much of anything, really. At least, I don't see a way to get past that problem.
Now, I'll give you my argument for this. Define the following set:
$$E = \{a_n: n \in \mathbb{N} \}$$
Since $\{a_n\}$ converges, it is bounded. So, $E \neq \varnothing$ and is bounded. Hence, $sup(E)$ exists and $inf(E)$ exists. I will show that:
$$a \leq sup(E)$$
Suppose that $a > sup(E)$. Then, there is an $\epsilon > 0$ such that:
$$a-\epsilon > sup(E)$$
So, there exists an $N \in \mathbb{N}$ such that:
$$\forall n \geq N: a_n \in (a-\epsilon,a+\epsilon)$$
This implies that:
$$\forall n \geq N: a_n > a-\epsilon > sup(E)$$
which is a contradiction. So, it follows that $a \leq sup(E)$. A similar argument can be made for the infimum.