$I$ is closed in the subspace topology.

The question you have asked presents a basic knowledge of topology---you are talking about open and closed sets. If you lack the topological foundation to understand that $I$ is a closed set when viewed as a subset of itself, then it isn't really going to be possible to answer the question you have asked in a manner which is going to be useful to you. But you can fall back on the $\varepsilon$-$\delta$ definition of continuity, and be just fine.

@XanderHenderson I shouldn't really be participating in this conversation (as I know nothing about topology), but doesn't $\{c\}$ also satisfy the definition for being an open Euclidean space?

@SohamSaha Who said anything about "open Euclidean spaces"?

@XanderHenderson the post mentions the set $\{c\}$ to be closed, but isn't it open if seen as a subset of $\mathbb{R}$ ?

@SohamSaha No. Singleton sets are closed in the standard topology on $\mathbb{R}$.

@XanderHenderson I am quite confused now. Doesn't the singleton set satisfy the definition in the link? Or is the definition there lacking something?

@SohamSaha No, it does not. If $U$ is open, then for any $x \in U$, there must be some $\varepsilon > 0$ such that $|x-y| < \varepsilon$ implies that $y \in U$. In the case of a singleton set, no such $\varepsilon$ exists for the single point in the set.

@XanderHenderson then is it to be assumed that $x\neq y$ in that case?

@SohamSaha No, but that question doesn't make sense to me. If $x$ is an element of an open set, then so to are all of the points within some $\varepsilon$ of $x$. This implies that, at the very least, there must be some $\varepsilon > 0$ such that the interval $(x-\varepsilon, x+\varepsilon)$ is a subset of any open set which contains $x$. Singleton sets do not contain any intervals, so they can't be open.

Yes? @SohamSaha

All I am asking is does the singleton set satisfy the definition given in the 1st line of "Euclidean Space" part?

You mean this:

Huh... the image uploader isn't working...

Extremely sorry for having wasted your time, but I think I have got it now...

In any event, no. A singleton does not meet that definition.

The definition means that for every x in (x-e,x+e) should be in U, right?

@SohamSaha Almost. For every $x \in U$, there must be some $\varepsilon$ such that $(x-\varepsilon, x+\varepsilon) \subseteq U$.

So a more intuitive approach might be that there exists a small circle around x, for every x in U, so that every element in that circle is in U right?

Sure, though don't use the term "circle". A circle is a set of points which are exactly a given distance from the center. You should use the term "ball", which is the set of all points which are less than some fixed distance from the center.

In one dimension, a ball is an interval. In two dimensions, a ball is a disk. In higher dimensions, just call it a ball.

Agreed and understood.

Thanks a lot for clarifying.

Should I delete my comments on the original post?

You can try...

...if there remain any comments to delete, which apparently don't

I have some basic knowledge about this topic, any material I can refer to for more insight? Does this fall entirely under topology?

@XanderHenderson ?

Read munkres