How to prove that if $a$ belongs to $\mathbb R$, such that $0\leq a \leq\epsilon$, then $a = 0$

I am taking a real analysis course. I have the following statement:

Prove that if $a$ belongs to $\mathbb R$, such that $0\leq a < \epsilon$, for all $\epsilon > 0$, then $a = 0$

I know how to prove it (using contradiction). Now, I was wondering, how I prove almost the same thing:

Prove that if $a$ belongs to $\mathbb R$, such that $0\leq a \leq \epsilon$ for all $\epsilon > 0$, then $a = 0$

Can someone show me how to do it? I am stuck with the equality to show a contradiction.

Note: this is not homework, just for my own knowledge.

If $a>0$ then $0\leq a\leq \epsilon$ is not true for $\epsilon=\frac{a}{2}>0$

If $a<0$ then it is immediate that $0\leq a\leq \epsilon$ is not true.

So the only 'candidate' that remains is $a=0$ and it is obvious that the statement is true for $a=0$

If this is the situation Najid Idrissi described it as this another way to prove it.

Assume $a\in\mathbb{R}$ such that for $\epsilon>0$ that $0\leq a\leq\epsilon$ (i.e. $a\in[0,\epsilon]$). Well this means that $a\in[0,\frac{1,}{n}]$ for $n>0$. This then indicates that $$a\in\bigcap_{n=1}^{\infty}\left[0,\frac{1}{n}\right]$$ Well we note that $$0\in\bigcap_{n=1}^{\infty}\left[0,\frac{1}{n}\right]$$ Thus all we must prove is that there is no other element within intersection. Seeking a contradiction suppose that there is a $c\neq0$ such that $c\in\bigcap_{n=1}^{\infty}\left[0,\frac{1}{n}\right]$. Well since $c\neq0$ then $c>0$ but this means that $c\not\in[0,\frac{c}{2}]$ but this means that for $m\in\mathbb{N}$ such that $m>\frac{2}{c}$ then for all such $m$ $c\not\in[0,\frac{1}{m}]\subset[0,\frac{c}{2}]$ which is a contradiction thus $$\bigcap_{n=1}^{\infty}\left[0,\frac{1}{n}\right]=\{0\}$$ and since $a\in\bigcap_{n=1}^{\infty}\left[0,\frac{1}{n}\right]$, we have that $a\in\{0\}$ thus $a=0$