$\{a_n\} \to a$ iff $\limsup_{n \to \infty} \{a_n\} = \liminf_{n \to \infty} \{a_n\}$ It is clear that if $$\limsup_{n \to \infty} \{a_n\} = \liminf_{\to \infty} \{a_n\},$$ then $\{a_n\} \to a$, since we can just squeeze the terms in the middle.   
I understand that to prove the opposite direction, it is enough to use the fact that all subsequences converge. As a means of trying to prove things more then one way, I'd like to offer a possible proof by contradiction.
Suppose $\{a_n\} \to a$. Let $E$ be the range of the $\{a_n\}$. Then $E$ is bounded. Now, it suffices to show that if $$\liminf_{n \to \infty} \{a_n\} \neq a,$$ then we arrive at a contradiction.
Let $\epsilon_2> \epsilon_1 > 0$. Let $\epsilon_2$ be such that, for all $n,N \in \mathbb{N}$, $$| \sup_{N} \inf_{n \geq N} \{a_n\} -a | > \epsilon_2.$$ Let $s_n = \inf \limits_{n \geq N} \{a_n\}$. For each $s_n$, by definition, there exists a subsequence of $\{a_n\}$ converging to $s_n$. Let $S_N = \sup \limits_{N} \inf \limits_{n \geq N} \{a_n\}$. By properties of $\sup$, there exists a subsequence of $\{s_n\} \to S_N$. Since the set of sub-sequential limits is closed, this means there exists a subsequence of $\{a_n\} \to S_N$. Take $M \in \mathbb{N}$ so that, for $m > M$, $$|a_m - a| < \epsilon_2-\epsilon_1.$$ Then, consider $S_M$. There exists a subsequence $\{a_{n_j}\} \to S_M$. In particular, $\exists J \in \mathbb{N}$ so that, for $n_j > J$, $$|a_{n_j} - S_M| < \epsilon_1.$$
Now, we have, for $k > \max\{M,J\}$, $$|a_k -a| \geq |a_{k} - S_M| - |S_M - a| > \epsilon_2 - \epsilon_1,$$ a contradiction. So, we're done.
Any issues?
 A: 
As a means of trying to prove things more then one way

Good idea; I would recommend trying to prove it by translating the $\epsilon$-$N$ definitions alone, without considering subsequences. Try not to rely on proof by contradiction.

Let $\epsilon_2> \epsilon_1 > 0$. Let $\epsilon_2$ be such that...

That's a confusing way of putting it. At first it sounds like $\epsilon_2$ is arbitrary, and then you add another condition to it. Try describing up front what the role of your variables is. Why do we need these epsilons, anyway? What's their tactical purpose?

$\sup_{n} \inf_{n \geq N} \{a_n\}$

This notation doesn't really mean anything, since $\inf_{n \geq N} \{a_n\}$ is not a function of $n$. The $\inf$ symbol has "captured" that variable. Perhaps you meant $\sup_{N} \inf_{n \geq N} \{a_n\}$. But then, both $n$ and $N$ have been captured, so it doesn't make sense to say "for all $n,N \in \mathbb{N}$". Again, try to be clearer about what your variables mean and what constraints apply to them.
A: The concept is okay, but there are a number of issues with how you've written it up:

Let $ϵ_2>ϵ_1>0$. Let $ϵ_2$ be such that, for all $n,N\in \Bbb N,
|\sup\limits_n \inf\limits_{n≥N} a_n−a|>ϵ_2$.

First, as stated you are introducing $\epsilon_2$ twice. You should say "let $\epsilon_2$ be such that ..., and let $ϵ_2>ϵ_1>0$", the latter relation be a restriction on the newly introduced $\epsilon_1$ only.
More importantly, and this problem runs all through your proof: $\inf\limits_{n≥N} f(n)$ is an expression that depends on the variable $N$, but not the variable $n$, which only appears in it as a dummy variable. So you would need to take the "$\sup\limits_N$", not "$\sup\limits_n$", of it. And having taken that supremum, $N$ is now a dummy variable as well. The expression $|\sup\limits_N \inf\limits_{n≥N} a_n−a|$ does not depend on $n$ or $N$.

Let $s_n = \inf \limits_{n \geq N} \{a_n\}$. For each $s_n$, by definition, there exists a subsequence of $\{a_n\}$ converging to $s_n$.

Again, that should be $s_N$. Also, this isn't quite by definition, but it is trivial enough you don't need to do more than state it.

Let $S_N = \sup \limits_{n} \inf \limits_{n \geq N} \{a_n\}$.

This value, which is the lim inf of $\{a_n\}$, does not depend on N. There is only one such value $S$.

Since the set of sub-sequential limits is closed,

Try not to make people scratch their heads and guess what it is you mean! "$\{s_n\}$ is a subsequence of $\{a_n\}$, and therefore any subsequence of $\{s_n\}$ is also a subsequence of $\{a_n\}$" would be better.
Since $S$ does not depend on $M$, the whole introduction of $M$ should be removed. You already had what you needed here.
And the final issue is that Chris Culter can type faster than I can!
