For the set $S$={x$\in$[a,b]: f(x)$\leq$y} where f is continuous, and f(a)< y < f(b),
I am trying to show that f($\alpha$=$Sup{S}$)=y.
So suppose not.
If f($\alpha$)>y, and since $\forall$$\epsilon$>0, $\alpha$-$\epsilon$ is not an upper bound, then f($\alpha$-$\epsilon$)< y < f($\alpha$), with equality not holding due to the preceeding assumption.
Although it might be irrelevant to the discussion of the IVT, my main question is; by using the quantifier $\forall$$\epsilon$>0, do I create a contradiction by showing that y is always between these two numbers, f($\alpha-\epsilon$) and f($\alpha$), even though the LHS converges to the RHS with choice of $\epsilon$?
Additionally, in what other ways can I "epsilon squeeze" an inequality in my proofs for future reference?
You need to fix your argument a bit. What is guaranteed is not that $f(\alpha-\epsilon) <y$ but rather that there is a member $\alpha'\in S$ such that $\alpha-\epsilon<\alpha'\leq \alpha$ and thus $f(\alpha') \leq y$.
To write the proper proof we don't need to consider all $\epsilon>0$. We just choose $\epsilon=1/n,n\in\mathbb {N} $ and thus for every $n\in\mathbb{N} $ we have an $\alpha_n\in S$ such that $\alpha-(1/n)<\alpha_n\leq \alpha $. And by Squeeze Theorem we have $\alpha_n\to\alpha$ and since $f(\alpha_n) \leq y<f(\alpha) $ taking limits as $n\to\infty$ and noting the continuity of $f$ we get $f(\alpha) \leq y<f(\alpha) $ so that a clear contradiction is obtained. A similar argument can be provided when $f(\alpha) <y$.