For $\epsilon > 0$ show that there exists $n_{0}$ so that $d_{H}(E_{n},E_{m}) < \epsilon$ for $n, m \geq n_{0}$ where $E_{n}$ is the graph $$E_{n} = \{(t,t^{n}) \; \colon \; t \in [0,1]\}.$$ Hint: Perhaps it is easier to spot a set $E$ with $d_{H}(E_{n},E) < \epsilon$ for large $n$ and use the triangle inequality
My question is should the interval in the definition of $E_{n}$ be $[0,1)$ instead? An obvious candidate for $E$ from the hint is the set $E = \{(t,0)\; \colon \; t \in [0,1)\}$ if this were the case. The question doesn't specify, but I believe we are working with $t \in \mathbb{R}$ as well. Since we can always find $n$-th roots in the real numbers, isn't the set $E_{n}$ as it's defined above the same for all $n$? Some clarification would be great, or perhaps I'm completely wrong and you can point me in the right direction.
The metric stated is the Hausdorff metric.
I think the problem is stated correctly and that the hint is good. The set $E$ does not need to be the graph of a function but it does need to be compact in order to apply the Haudorff metric. Indeed, I think a major point behind the problem is to help you distinguish between closeness in the Hausdorff metric and closeness in the supremum norm, which is really just defined between functions.
For the set $E$, I would choose $$E = \{(s,0): 0\leq s \leq 1\} \cup \{(1,t): 0 \leq t \leq 1\}.$$ This is the union of a horizontal line segment with a vertical line segment, as shown in the image below. The image also shows the graph of the function $f(x)=x^{80}$.
As we see, every point on the graph is close to some point in the set $E$ and vice-versa. Intuitively, this is what we mean for two sets to be close in the Hausdorff metric and I think it's fairly straighforward to formalize a proof from here.