Rigor in proving continuity of $f$ over a closed interval $I$ Given a function $f$ on a closed interval $I \subset \mathbb{R}$, where $I = [a,b]$, to prove continuity of $f$ over the interval $I$, what is generally done is the following.
1. We prove that $f$ is continuous at endpoint $a$
$$ \lim_{x \ \to \  a^{+}} f(x) = f(a)$$
2. We prove that $f$ is continuous $\forall c \in (a, b)$
$$ \lim_{x \to c} f(x) = f(c) \ \ \ \ \text{,}\ \ \ \ \forall c \in (a, b)$$
3. We prove that $f$ is continuousat endpoint $b$
$$ \lim_{x \ \to \  b^{-}} f(x) = f(b)$$

My question revolves around step 2 of this process. Usually all that is done to prove the continuity of $f$ over the open interval $(a, b)$, is (using the Limit Laws) a simple direct substitution of a variable $c$ denoting an arbitrary number in the open interval, which is basically substituting $c \in (a, b)$ into $f(x)$.
But I don't see why this is mathematically rigorous? Initially I assumed something along the lines of Mathematical Induction would have been used to prove $f$ to be continuous $\forall c \in (a, b)$, but that doesn't seem to be the case. 
Why does just choosing an arbitrary $c$ to represent a number in the interval $(a, b)$, and showing that the chosen $c$ satisfies the given definition of continuity, prove that $f$ is continuous over $(a, b)$?
I will give an example below to highlight what I'm asking.

Example : Prove $f(x) = e^{x}$ to be continuous over the open interval $(0, 2)$
Proof : Let $c \in (0, 2)$
$$\begin{equation}
\begin{aligned}
\lim_{x \to c} f(x) &= \lim_{x \to c} e^{x} \\
&= e^{c} \\
&= f(c)\\
\end{aligned}
\end{equation}$$
$$Q.E.D$$
(This is how most proofs for continuity of functions over intervals are treated (at least in first-year Calculus courses). But this way of proving continuity doesn't seem rigorous to me at all. It does nothing to explicitly state or show that $c$ is an element of the open interval $(0, 2)$ other than the fact that we assume it to be true. I could say 'Let $c \in \mathbb{R}$', and the proof would still hold, even though $f(x)$ is only defined for $c \in \mathbb{R^{+}}, \ \text{where} \ c \neq 0$.

Just stating that $c \in (a, b)$ and showing that $lim_{x \to c} f(x) = f(c)$, doesn't seem like it proves anything to me, because $c$ doesn't necessarily need to be bounded within that interval, as we have done nothing to explicitly show $c \in (a, b)$ other than the fact that we stated it or assumed it. 
I mean using this sort of proof process, I could say let $c$ be an element of any arbitrary interval $R$, and by simply using direct substitution to show that $lim_{x \to c} f(x) = f(c)$, I would show that $f$ would be continuous over $R$, even if $f$ contained discontinuities over $R$, which this sort of proof process doesn't take into account at all.
To re-emphazize, what I'm getting at, what I'm trying to say is I could always pick an arbitrary $c$ in any arbitrary open interval $I$, and using the Limit Laws (as used in this way in these sorts of proofs) always prove $f$ to be continuous on $I$ even if  $I \not\subset D$, i.e. the open interval lies outside of the domain of the function.
As a last sub-question, in Real Analysis is there a more rigorous way in which continuity of a function $f$ over a closed interval $I$ is proven? Is there a reason in Real Analysis why continuity of functions over closed intervals is proved in this sort of way? Essentially I'm looking for a deeper understanding of proving continuity of functions over intervals, and I'm sure Real Analysis must have the answers to the questions I'm asking.
 A: I don't see any problem choosing a generic point $c \in I$ to check continuity on $I$ because continuity on an interval in defined in that manner. If the steps involved in proving continuity at $c$ also hold if $c$ lies outside $I$ then it is not a problem (your function is continuous outside of $I$ also unless you specifically mention that $f$ is defined only on $I$ and not elsewhere). What is important is to ensure that the steps don't fail for any $c \in I$. Unless the behavior of a function is not highly specific for some specific point we can chose a generic $c \in I$.
It is worthwhile to consider the simple example $f(x) = x\sin(1/x)$ for all $x \neq 0$ and $f(0) = 0$. This means that $f$ is defined for all $x$. I will prove that it is continuous on $\mathbb{R}$. Let $c \in \mathbb{R}$ and then $$\lim_{x \to c}f(x) = \lim_{x \to c}x\sin(1/x) = c\sin (1/c) = f(c)\tag{1}$$ and hence $f$ is continuous on $\mathbb{R}$. This is wrong.
The equation $(1)$ is not valid when $c = 0$ but at the same time it is valid if $c \neq 0$. Hence using equation $(1)$ you can prove continuity for all $c \in \mathbb{R}$ except $c = 0$. But for $c = 0$ you need to do something different. Moreover we need to understand why $(1)$ is valid for any $c \neq 0$. The reason is that the function $g(x) = \sin x$ is continuous for all $x \in \mathbb{R}$ (this requires a proof and it is not difficult) and the function $h(x) = 1/x$ is continuous for all $x \neq 0$ (this is proven using limit laws). Now using these facts and limit laws each step in $(1)$ is justified as long as $c \neq 0$.
What to do for $c = 0$? Well we have $$0 \leq |f(x)| \leq |x|$$ for all $x \neq 0$ (the inequality is true for $x = 0$ also, but it does not matter here). Hence letting $x \to 0$ and using Squeeze theorem we get $$\lim_{x \to 0}|f(x)| = 0$$ This implies (why? prove it via definition of limit) that $$\lim_{x \to 0}f(x) = 0 = f(0)$$ so that $f$ is continuous at $0$ also.
