Prove that $\bigcap_{x\in[-1,0]}[x,1+x] = \{0\}$ $$\bigcap_{x\in[-1,0]}[x,1+x]=\{0\}$$
My try:
So my approach is a little bit different than the usual, that $\{0\}\subset \bigcap_{x\in[-1,0]}[x,1+x]$ and that $\bigcap_{x\in[-1,0]}[x,1+x]\subset {0}$$.
So I tried to do it that we have $a\in[x,1+x],$ $x\in[-1,0]\implies a\in [-1,1]$, and is a real number.
So now if we look at $x = -1+ \epsilon$, where $\epsilon>0$ and is a small real number, we have some $a_0 \in [-1+\epsilon,\epsilon]$ and if we look at $x=-\epsilon$, we have $a_1\in[-\epsilon,1-\epsilon]$.
Ok, so what I want to show is that if $a_1,a_0\in \bigcap_{x\in[-1,0]}[x,1+x]$ iff $a_1 =a_0$.
Now if we look at the inequalities (we say that $\{a \in \mathbb{R}|\forall x \in [-1,0]:a \in \bigcap[x,1+x]\}$):
$$-1+\epsilon \leq a \leq \epsilon \quad \wedge \quad -\epsilon \leq a \leq 1-\epsilon$$
$$-\epsilon \leq a \leq \epsilon$$
and as $\epsilon \rightarrow 0$, we get: $0\leq a\leq0 \implies a = 0 = a_1 = a_0$
Is my proof correct? How would you show it the usual way (with subset going both ways)?

Your approach shows $\bigcap_{x \in [-1,0]} [x,1+x] \subseteq \{0\}$. It could be also written like this: let $a \in \bigcap_{x \in [-1,0]} [x,1+x]$ be arbitrary. Then for any $x \in [-1,0]$ we have
$$x \le a \le 1+x.$$
Letting $x = 0$ we get $0 \le a$ and letting $x = 1$ we get $a \le 0$. Therefore $a = 0$.
The converse inclusion is easy. For any $x \in [-1,0]$ we have $1+x \in [0,1]$ and hence $$x \le 0 \le 1+x \implies 0 \in [x,1+x].$$ Since $x \in [-1,0]$ was arbitrary, we conclude
$$0 \in \bigcap_{x \in [-1,0]} [x,1+x] \implies \{0\} \subseteq \bigcap_{x \in [-1,0]} [x,1+x].$$
