Let $f : I \rightarrow \mathbb{R}, \space f \space$ is continuous and has the following property:
$\forall \space (a, b) \subset I, \space \exists \space c \in (a, b) \space$ so that $f(c) = 0$.
Show that $\forall \space x \in I, \space f(x) = 0$.
Proof:
Let $y \in I$
Since $f \space$ is continuous this means $$\lim_{x \rightarrow \space y}f(x) = f(y)$$
From the definition $\space(\delta, \varepsilon)\space$ of limits of functions this implies that $\forall \space \varepsilon > 0, \space \exists \space \delta(\varepsilon) = \delta > 0, \space$
so that if $\space |f(x) - f(y)| < \varepsilon \space$ then $|x - y| < \delta$
Assume $f(y) > 0$ and let $\varepsilon \le f(y) \Longleftrightarrow \space 0 \le f(y) - \varepsilon$
\begin{align*} &|f(x) - f(y)| < \varepsilon \\ & 0 \le f(y) - \varepsilon < f(x) < \varepsilon + f(y) \\ & 0 < f(x) < \varepsilon + f(y) \end{align*}
This concludes to $\forall \space \varepsilon \in (0, f(y)), \space f(x) \ne 0$.
Because $|x - y| < \delta$, then $y - \delta < x < y + \delta$ so this means that $x \in (y - \delta, y + \delta)$ and because $ f(x) \ne 0$, let $a = y - \delta$ and $b = y + \delta$ $\space \nexists \space c \in (a, b), \space f(c) = 0$, this is a contradiction.
The case when $f(y) < 0$ also leads to a contradiction, this can be similarly proved by taking $\mathcal{E} \le -f(y)$.
This ends with the third case when $f(y) = 0$, since $y$ was arbitrarily chosen this means that $\forall \space x \in I, \space f(x) = 0$.
You have the definition of limit backwards. The correct one is
This makes the proof invalid, I'm afraid.
Probably your assumption is that $I$ is an open interval.
The proof is then much easier. Suppose $f(y)>0$. Then there exists $\delta>0$ such that, for $|x-y|<\delta$, $x\in I$, $|f(x)-f(y)|<f(y)/2$, which implies $f(x)>f(y)/2$.
Since you can choose a smaller $\delta_1>0$ such that $(y-\delta_1,y+\delta_1)\subset I$, this contradicts the assumption.
Similarly if $f(y)<0$.