Question: Say the topological space $T$ having a basis as the $B=\{(a,b] \vert a,b\in \mathbb{R}, a<b\}$on the real number set, $\mathbb{R}$. Defining the $f\_1, f\_2 : \mathbb{R} \to (\mathbb{R},T)$ by $f\_1(x) = x^2$ and $f\_2(x) = -x^2$. Find the closure of the $(0,1]$ in the $(\mathbb{R}, T\_1)$. Here the $ T\_1$ is the minimal topological space such that both $f\_1, f\_2$ are continuous.

From now, My solution starts.
Try to solve the above question, I find the form of the basis that $f\_1^{-1}(G) \cap f\_2^{-1}(H)$ for the $G,H\in (\mathbb{R},T)$.
Hence, The minimal open set of the each points are $\{\pm a\}, (a\neq 0)$ or $(-\epsilon, \epsilon)$. So my answer is $[-1,1]$ considering the accumulation points. Is my answer right?

Answer: 
I think your answer is not correct. First of all $[-1,1]$ is closed in your topology because

$\mathbb{R}\setminus[-1,1]=f\_1^{-1}(1,\infty)$

Thus the closure $C$ of $(0,1]$ is contained in $[-1, 1]$. Let us suppose there is $a\in [-1,1]\setminus \{0\}$ that is not in $C$ (so $a<0$).

Any open neighbourhood of $a$ contains $f\_1^{-1}(-\epsilon+a^2, a^2]$ that intersect always $(0,1]$ in $-a$. This means that $a$ is a point in the closure of $(0,1]$ and so $a\in C$, that is a contradiction.

Moreover $\{0\}$ is not in the clousure of $(0,1]$ because $\{0\} =f\_1^{-1}((-1,0]))$ does not intersect $(0,1]$.

Thus $C=[-1,1]\setminus \{0\} $.

You can do a similar reasoning for the map $f\_2$
Notice that, in the case $0\le a$, $f\_1^{-1}(a,b]=(\sqrt{a},\sqrt{b}]\sqcup[-\sqrt{b},-\sqrt{a})$. We certainly do not have $\{a\}$ - your mistake there probably came from "$f\_1^{-1}(-a^2,a^2])=\{a\}$" but this is false since $f\_1^{-1}(-a^2,0]=\{0\}$ and $f\_1^{-1}(0,a^2]=(0,a]\sqcup[-a,0)$, and the union thereof is $[-a,a]$. So we have $[-c,c]$ and $(c,d]\sqcup[-d,-c)$ as the basis sets derived from $f^{-1}\_1$, with $c,d\ge0$.

For $f\_2$, we have similar situations. $f\_2^{-1}(-a^2,a^2]=f\_2^{-1}(-a^2,0]=(-a,a)$, and $f\_2^{-1}(a,b]$ where $b\le 0$ yields $[\sqrt{-b},\sqrt{-a})\sqcup(-\sqrt{-a},-\sqrt{-b}]$. So the basis sets derived from $f\_2^{-1}$ are $(-c,c)$ and $[c,d)\sqcup(-d,-c]$ for $c,d\ge0$.

Now let's take the closure of $(0,1]$ using these basis sets. The first question to ask is: can an $x\gt1$ be a boundary point of this set? No, it cannot, as I can consider the open neighbourhood $(-x-1,-x]\cup[x,x+1)$ which is disjoint from $(0,1]$. Can $x=0$ be a boundary point? No, as $\{0\}$ was shown to be open. Can $-1\le x\lt0$ be a boundary point? Since every basis set is symmetric, a neighbourhood containing such an $x$ would also hold $-x\in(0,1]$. Finally, can $x\lt-1$ be a boundary point? No, for the same reasons as $x\gt1$ cannot be a boundary point.

So, $[-1,1]\setminus\{0\}$ is correct and your basis sets were not right.