In the First Fundamental Theorem of Calculus, why must $F : [a, b] \rightarrow \mathbb{R}$ be continuous at the interval endpoints? 
Show that the First Fundamental Theorem (Integrating Derivatives) it is necessary to assume that the function $F : [a, b] \rightarrow \mathbb{R}$ is continuous at the endpoints of the interval.

I'm aware that in the last step of the proof, when the Mean Value Theorem is applied on each subinterval $[x_{i - 1}, x_{i}]$ for $i\geq1$, it is required that $F(x)$ is continuous at each subinterval $[x_{i - 1}, x_{i}]$ for $i \geq 1$, so taking $x_{0} = a$ and $x_{n} = b$ suffices.  But does this explanation really suffice for this question? Since the question says "show," I think it would be better for me to show where it breaks down, but I haven't been able to do that. 
For reference, the First Fundamental Theorem states that if the function $F' : (a, b) \rightarrow \mathbb{R}$ is both continuous and bounded, then, 
$$\int_{a}^{b} F'(x) \mathop{dx} = F(b) - F(a)$$
 A: You need to understand that a precondition for defining Riemann integral $\int_{a} ^{b} f(x) \, dx$ is that $f$ is defined and bounded on $[a, b] $. Thus for $\int_{a} ^{b} F'(x) \, dx$ to make any sense we must have $F$ differentiable on $[a, b] $ and therefore $F$ is continuous on $[a, b] $ so that continuity is not an assumption but rather an obvious conclusion.
It is rather weird that you have an exercise to show the necessity of this assumption.

After your comments to the question, it appears that you are interested in the following question :

Let $F$ be defined on $[a, b] $ and differentiable on $(a, b) $ such that $F'$ is bounded and continuous on $(a, b) $. Do we necessarily need $F$ to be continuous on $[a, b] $ so that $$\int_{a} ^{b} F'(x) \, dx=F(b) - F(a)? $$

As I mentioned in first part of the answer you need some values of the function $F'$ at $a, b$ before we can make sense of its integral on $[a, b] $. Since the values of a function at a finite number of points does not impact its Riemann integrability or the the value of its integral (if it exists) we may as well consider a function $g:[a, b] \to\mathbb{R} $ such that $g(x) =F'(x) $ for all $x\in(a, b) $ and by the given conditions of $F'$ we see that $g$ is bounded and Riemann integrable on $[a, b] $ and thus the symbol $\int_{a}^{b} F'(x) \, dx$ can be defined as $\int_{a}^{b} g(x) \, dx$. 
Let's now consider the function $G:[a, b] \to\mathbb{R} $ defined by $$G(x) =\int_{a} ^{x} g(t) \, dt$$ And then by definition of $G$ we have $G$ continous on $[a, b] $ and further $$\int_{a} ^{b} g(x) \, dx=G(b) - G(a),G'(x)=g(x)=F'(x)\, \forall x\in(a, b) $$ Consider the function $H(x) = F(x) - G(x) $ then $H'(x) =0$ for all $x\in(a, b) $. It follows that $H(x) $ equals some constant on $k$ on $(a, b) $ and thus $$F(x) =G(x) +k\, \forall x\in(a, b) $$ It follows that $$\lim_{x\to a^{+}} F(x) =G(a) +k, \lim_{x\to b^{-}} F(x) =G(b) +k$$ and putting all pieces together we have $$\int_{a} ^{b} F'(x) \, dx=\int_{a} ^{b} g(x) \, dx=G(b) - G(a) =\lim_{x\to b^{-}} F(x) - \lim_{x\to a^{+}} F(x) $$ and thus continuity of $F$ at $a, b$ is not necessary but we only need to pick values $F(a), F(b) $ such that $F(b) - F(a) $ equals the difference between left hand limit of $F$ at $b$ and right hand limit of $F$ at $a$. 
I hope this is what you intend to know, but this is not exactly clarified in your question.

The argument of my answer can be simplified greatly if one considers a fixed point $c\in(a, b) $ and then note that by FTC we have $$F(x) =F(c) +\int_{c} ^{x} F'(t) \, dt, \,\forall x\in(a, b) $$ Since the integral $\int_{a} ^{b} F'(x) \, dx$ exists it follows by above equation that left (right) hand limit of $F$ at $b$ ($a$) exists and note that the above argument works for not just continuous derivative $F'$ but rather for any derivative $F'$ which is bounded and Riemann integrable. 
A: [Updated: 11/8/18]

[Statement A from OP] For reference, the First Fundamental Theorem states that if the function $F' : (a, b) \rightarrow \mathbb{R}$ is both continuous and bounded, then,
$$
\int_{a}^{b} F'(x) \mathop{dx} = F(b) - F(a)\tag{1}
$$

This cited "First Fundamental Theorem" is incorrect. Consider the following simple counterexample. Let $F:[0,1]\to{\mathbb R}$ be such that $F(x)=0$ on $[0,1)$ and $F(1)=1$. Note that $F':(0,1)\to{\mathbb R}$ is both continuous and bounded but
$$
\int_0^1F(x)\ dx=0\neq F(1)-F(0).
$$ 


*

*Now, in order to turn A into a correct statement, must one assume that $F$ is continuous at both the endpoints $a$ and $b$?
No, since one can simply consider the example of $F:[0,1]\to{\mathbb R}$ where $F(x)=0$ for $x\in(0,1)$ and $F(0)=F(1)=1$. 

*What assumption of $F$ can one add to the statement A so that one can conclude (1)? 
Theorem 7.34 (Second fundamental theorem of integral calculus) in Apostol's Mathematical Analysis (2nd edition 1974, page 162) states the following:

Assume that $f\in{\mathbb R}$ on $[a,b]$. Let $g$ be a function defined on $[a,b]$ such that the derivative $g'$ exists in $(a,b)$ and has the value
  $$
g'(x)=f(x)\quad \forall x\in(a,b).
$$
  At the endpoints assume that $g(a+)$ and $g(b-)$ exist and satisfy
  $$
g(a)-g(a+)=g(b)-g(b-).
$$
  Then we have
  $$
\int_a^bf(x)\ dx = \int_a^b g'(x)\ dx= g(b)-g(a).
$$


*Also, see Paramanand Singh's answer. 

[The following original answer is based on an "incorrect" assumption that the statement A cited by OP is correct, which is not.]
Your cited version of FTC is usually called the second fundamental theorem of calculus in references. 
For necessity, note that your version of FTC implies that
$$
\int_a^xF'(t)\ dt=F(x)-F(a),\quad x\in [a,b],
$$
equivalently,
$$
F(x)=F(a)+\int_a^xF'(t)\ dt\quad x\in [a,b].
$$
Since the map
$$
x\mapsto \int_a^xF'(t)\ dt
$$
is continuous on $[a,b]$ by the first fundamental theorem of calculus, $F$ must be continuous on $[a,b]$ as well. In particular, $F$ is continuous at both $x=a$ and $x=b$.
