Using the $\varepsilon$ definition of interior to show that $S=\{x\in\mathbb{R}^2:0My thoughts:
By def. of interior
\begin{align}
S^\circ
&=\{x∈\mathbb{R}^2:∃ε>0,B(x,ε)⊆S\}\\
&=\{x∈\mathbb{R}^2:∃ε>0,\forall t\in\mathbb{R}^2,\Vert t-x\Vert<\varepsilon\rightarrow(0<t_1<1\land0<t_2<1)\}
\end{align}
WTS $S=S^\circ$ that is
\begin{align}
&\{x\in\mathbb{R}^2:0<x_1<1\land0<x_2<1\}\\
&=\{x∈\mathbb{R}^2:∃ε>0,\forall t\in\mathbb{R}^2,\Vert t-x\Vert<\varepsilon\rightarrow(0<t_1<1\land0<t_2<1)\}
\end{align}
And we want to prove that any $x\in\mathbb{R}^2,x\in S\text{ iff } x\in S^\circ$
That is same as 
$$\forall x\in\mathbb{R}^2,\big(0<x_1<1\land0<x_2<1$$
$$\text{iff }∃ε>0,\forall t\in\mathbb{R}^2,\Vert t-x\Vert<\varepsilon\rightarrow(0<t_1<1\land0<t_2<1)\big)$$
My Atempts:
$\Leftarrow:$
It is trivial, after we varify that $\exists x\in\mathbb{R}^2,\exists\varepsilon>0,\exists t\in\mathbb{R}^2,\Vert t-x\Vert<\varepsilon$ is satisfiable
This direction will clearly just be a tautology.
$\Rightarrow:$
My intuition:

First Let $x\in\mathbb{R}^2$ such that
$$0<x_1<1\land0<x_2<1$$
Pick $\varepsilon=\min\{d\in\mathbb{R}:\forall y\in\partial S,d=\Vert x-y\Vert\}$
Let $t\in\mathbb{R}^2$ such that
$$\Vert t-x\Vert=\sqrt{|t_1-x_1|^2-|t_2-x_2|^2}<\varepsilon$$
Then the goal is to prove $$0<t_1<1\land0<t_2<1$$
Yet, I don't know how to prove this direction $\dots$
Any help would be appreciated.
 A: Given $x = (x_1, x_2)$ and
$\tag 1 0 \lt x_1 \lt1 \; \land \; 0 \lt x_2 \lt 1$
Set 
$\tag 2 r_1 = \min \big(|x_1|, |1-x_1|\big) \text{ and } r_2 = \min \big(|x_2|, |1-x_2|\big)$
Since the OP likes $\varepsilon$, set
$\tag 3 \varepsilon = \min \big(r_1,r_2\big)$
Let $t = (t_1, t_2)$ satisfy
$\tag 4 \sqrt{|t_1-x_1|^2+|t_2-x_2|^2}<\varepsilon$
You can now break things down into cases to come up with a proof. The two main case are
CASE 1: $\varepsilon = r_1$
...subcases
CASE 2: $\varepsilon = r_2$
...subcases
A: The OP stated in two comments that they wanted to work on the algebra. In this answer we state some general theory that can be proved using algebra/analysis.
The OP can employ this theory to solve their problem.

Assume the following constants:
$\quad a_1, b_1, a_2, b_2, \varepsilon \in \Bbb R$
Assume the following properties/relations:
$\quad \varepsilon \gt 0$
$\quad b_1 - a_1 = \varepsilon $
$\quad b_2 - a_2 = \varepsilon $
Define
$\quad x_1 = \frac{a_1+b_1}{2}$
$\quad x_2 = \frac{a_2+b_2}{2}$
$\quad r_1 = b_1 - x_1$
$\quad r_2 = b_2 - x_2$
Lemma 1: The number $r_1$ is equal to the number $r_2$.
We denote this number by $r$.
Lemma 2: The number $r$ is greater than $0$.
Proposition 3: For every $t_1,t_2 \in \Bbb R$ we have
$ (t_1-x_1)^2+(t_2-x_2)^2  \lt r^2 \text{ implies } \bigr [\, |t_1-x_1| \lt r \, \land \, |t_2-x_2|  \lt r \, \bigr ]$
A: Let $\varepsilon=\min(|1-x_1|,|x_1|,|1-x_2|,|x_2|)$. With this, it's easy to show that 
$\sqrt{(x_1-t_1)^2+(x_2-t_2)^2}<\varepsilon\rightarrow\\
\rightarrow \begin{cases}
|x_1-t_1|<\varepsilon\\
|x_2-t_2|<\varepsilon\end{cases}\rightarrow \begin{cases}
|x_1-t_1|<\min(|1-x_1|,|x_1|)\\
|x_2-t_2|<\min(|1-x_2|,|x_2|)\end{cases}\rightarrow \begin{cases}t_1\in (0,1)\\t_2 \in (0,1)\end{cases}$
(Note that the last step is justified noting that $x_1\in(0,1)\wedge x_2\in (0,1)$)
Geometrically, $\varepsilon$ is exactly the distance between $(x_1,x_2)$ and $\partial S$, as it is easy to see. Your intuition is thus effective
