Is there a proof that $\mathbb{R}$ is connected? Is there a proof that the set $\mathbb{R}$ of all real numbers is connected? I've been assuming that $\mathbb{Q}$ is discrete, with a (very small) gap existing between any two elements $x,y\in\mathbb{Q}$, and when you incorporate the irrationals you manage to "fill in" all such gaps. Is that actually how it works?
Edit: It sounds like I should probably be asking whether $\mathbb{R}$ is connected, not whether it's continuous. My line of thought is that while you can find an element $r\in\mathbb{Q}$ between any two $x,y\in\mathbb{Q}$, you can also find an element $s\notin \mathbb{Q}$ that lies between $x$ and $y$ when thinking in terms of their ordering on $\mathbb{R}$ (I realize that wording is kind of convoluted). Thus, $x$ and $y$ are disjoint, or "unconnected", which is what I meant by calling them discrete. I'm not sure that $s$ exists between any two $x$ and $y$, but it's my intuitive hunch.
I am a bit out of my depth, and could be misusing some of the terminology, so apologies for any confusion. I haven't studied topology yet, and I'm rusty on a lot of the advanced topics, which is why I'm reviewing some more basic stuff right now. As I read through familiar concepts, I keep finding myself questioning them in new light, and hypothesizing about stuff I'm not familiar with yet. Which is fun, and I'm looking forward to improving my understanding of these concepts.
 A: As it turns out, both $\Bbb R$ and $\Bbb Q$ are what is called a dense linear order--that is, a totally ordered set, such that between any two distinct elements, we can find a third. In that sense, neither of them has any gaps. The essential property that distinguishes $\Bbb R$ from $\Bbb Q$ is that $\Bbb R$ has the least upper bound property--any non-empty subset of $\Bbb R$ with an upper bound in $\Bbb R$ has a least upper bound in $\Bbb R$. A dense linear order with the least upper bound property is called a linear continuum. $\Bbb R$ is a linear continuum, but $\Bbb Q$ is not. That may be what you're trying to get to.
A: I think the notion you're looking for might be connectedness. If a set $S$ is connected, it means that you cannot write $S = A\sqcup B$ with $A$ and $B$ nonempty open subsets of $S$. Intuitively, if you try to separate a connected space into nonempty open subsets, there's going to have to be some overlap between them (try to take an open interval and write it as two disjoint open intervals!). And yes, there are proofs that $\Bbb{R}$ is connected (see here, for example). However, if you complete $\Bbb{Q}$ with respect to a different metric, you can obtain a space that is totally disconnected. So while your notion of "filling in the gaps" gives you the right idea for $\Bbb{R}$, that intuition doesn't carry over to a correct notion when taking other metric spaces (or even the same space with a different metric!).
Edit: The idea that there is some number in between any two rational numbers (when $\Bbb{Q}$ is taken with the topology induced the the standard absolute value, or the metric induced when $\Bbb{Q}$ is seen as a subset of $\Bbb{R}$) can be seen by taking an irrational number $\xi$ and looking at the two sets $I_1 = \left(-\infty,\xi\right)$, $I_2 = \left(\xi,\infty\right)$. It is not hard to see that these two sets cover $\Bbb{Q}$ and have empty intersection, so that $\Bbb{Q} = \left(I_1\cap\Bbb{Q}\right)\sqcup \left(I_2\cap\Bbb{Q}\right)$, and hence $\Bbb{Q}$ is disconnected, and "misses $\xi$." You can also perform this type of construction backwards, to separate any two rational numbers: let $p,q\in\Bbb{Q}$, and let $\xi\in\left[0,1\right]\setminus\Bbb{Q}$. Suppose $\left|p - q\right| = d$. We can find an irrational number $\xi'$ such that $p + \xi'<q$ by first finding an $n\in\Bbb{N}$ such that $10^{-n} \leq d$, and then replacing the first $n$ digits of the decimal expansion of $\xi$ with $0$'s: i.e., $\xi = 0.\xi_1\xi_2\xi_3\ldots\mapsto\xi' = 0.00\ldots\xi_{n+1}\xi_{n+2}\ldots$. $\xi$ Since a rational number plus an irrational number is irrational, there is an irrational number between any two rational numbers. To complete the process, let $\Bbb{Q} = \left(\left(-\infty, p + \xi'\right)\cap\Bbb{Q}\right)\sqcup\left(\left(p + \xi', \infty\right)\cap\Bbb{Q}\right)$.
In general, we can partition a topological space $X$ into connected components, which are maximal connected subsets with respect to inclusion. That is, if $\Gamma$ is a connected component of $X$, then $\Gamma\cup\{x\}$ (with $x\in X\setminus\Gamma$) will no longer be connected. What I did in the last paragraph actually shows that the connected components of $\left(\Bbb{Q},\left|\cdot\right|\right)$ are the singletons, which is the same as saying that $\left(\Bbb{Q},\left|\cdot\right|\right)$ is totally disconnected.
A: This is written for the student who has delved into 'Chapter 1' of mathematics and 'understands' the real number line. They have studied neither topological nor metric spaces, but accepts that both the $\text{lub}$ (respectively $\text{glb}$) exists when a subset of real numbers is bounded above (respectively below). Of course our student knows that the rational numbers contains 'gaps' such as the square root of $2$ and that the set $\{q \in \mathbb Q \; | \; q^2 \lt 2\}$ can't be expressed as an interval of rational numbers $(-\sigma,+\sigma)$ with $\sigma \gt 0 \text{ and } \sigma \in \mathbb Q$.
The following result provides some basic insight into why we regard the real numbers as 'gap-free' and 'complete'.
Proposition 1: Let $E$ be a subset of the real numbers $\mathbb R$ containing at least two numbers. Suppose the following two conditions hold:
$\tag 1 \text{IF } (x,y \in E) \text{ & } (x \lt y) \text{ & } (z \in \mathbb R) \text{ & } (x \lt z \lt y) \text{ THEN } z \in E$
$\tag 2 E \subset [-M,+M] \;\text{ for some positive real number } M$
Then there exist numbers $a,b \in \mathbb R$ such that $E$ is equal to one of the following four intervals:
$\quad \left( a, b \right)$
$\quad \left[ a, b \right)$
$\quad \left( a, b \right]$
$\quad \left[ a, b \right]$
Proof:
Set $b = \text{lub}(E)$ and $a = \text{glb}(E)$. $\quad \blacksquare$
We the leave the proof of the following theorem as a simple exercise.
Theorem 2: Let $E$ be a subset of the real numbers $\mathbb R$ satisfying $\text{(1)}$ and containing at least two numbers. Then there exist elements $a,b \in \mathbb R \cup \{-\infty, +\infty\}$  such that $E$ is equal to one of the following four intervals:
$\quad \left( a, b \right)$
$\quad \left[ a, b \right)$
$\quad \left( a, b \right]$
$\quad \left[ a, b \right]$

OK, now that our student has caught up and learned how the real numbers forms a topological space we can continue.
Recall that a topological space $X$ is said to be connected if $\emptyset$ and $X$ are the only subsets of $X$ that are both open and closed (called 'clopen' sets).
Theorem 3: The topological space $\mathbb R$ is connected.
Proof:
Start with any clopen $E$ and select a number $x \in E$. Since $E$ is open, there exists an open interval contained in $E$ having $x$ as an interior point. Consider all the open intervals $I$ of $\mathbb R$ contained in $E$ and with $x \in I$ . The union of these open intervals forms a set $I_x \subset E$ satisfying $\text{(1)}$. Applying theorem 2 we know that $I_x = (a,b)$ ($I_x$ is open in $\mathbb R$).
We claim that $a = -\infty$ and $b = +\infty$. Suppose $b \in \mathbb R$. Since $E$ is closed, $b \in E$. Since $E$ is open, there is an open interval containing $b$ that can be used to extend $I_x$, a maximal set. This makes no sense so $b = +\infty$ . An identical argument shows that $a = -\infty$. $\quad \blacksquare$
