Show that $\int_{x}^{x+c} f(t) dt$ is continuous 
Show that $F(x) =\int_{x}^{x+c} f(t) dt$ is continuous if $f$ is continuous, bounded. Is this true?

Here's my attempt:
$|F(x) - F(y)|  = |\int_{x}^{x+c} f(t) dt - \int_{y}^{y+c} f(t) dt|$. By a change of variables if $y = x+r$, then the equation above becomes $|\int_{x}^{x+c} f(t) - f(t-r) dt|$. Now for $\epsilon>0$ there is a $\delta$ such that since $|r| < \delta$, then $|f(t) - f(t-r)|< \epsilon$. Thus it is less than $c\epsilon$. Is my proof correct?
 A: It's not a bad attempt, but it needs some cleaning up.
First of all, I think you mean
$$\left|\int_x^{x + c} f(t) - f(t + r) \;\mathrm{d}t\right|.$$
Secondly, you also seem to be assuming uniform continuity. In general, the $\delta > 0$ such that $|r| < \delta \implies |f(t) - f(t + r)|$ will depend on the value of $t$. It might shrink as $t$ moves along the interval.
However, this is an easy fix. You can appeal to the fact that continuous functions on compact intervals are uniformly continuous. If you pick an interval $[a, b]$ containing $[x, x + c]$ and $[y, y + c]$, then $f$ is uniformly continuous on that interval, so the logic works.
Thirdly, I think you need to put the steps in the proper order. Start by assuming $\varepsilon > 0$. Using uniform continuity, you can find a $\delta > 0$, such that for all $t \in [x, x + c] \cup[y, y + c]$ such that
\begin{align*}
&|x - y| < \delta \implies |f(t) - f(t + y - x)| < \frac{\varepsilon}{2c} \\
\implies \; &\int_{x}^{x + c}|f(t) - f(t + y - x)| \; \mathrm{d}t \le \int_{x}^{x + c}\frac{\varepsilon}{2c} \; \mathrm{d}t \\
\implies \; &\left|\int_{x}^{x + c}f(t) - f(t + y - x) \; \mathrm{d}t\right| \le \frac{\varepsilon}{2} \\
\implies \; &\left|\int_{x}^{x + c}f(t) \; \mathrm{d}t - \int_{y}^{y + c}f(t) \; \mathrm{d}t \; \mathrm{d}t\right| < \varepsilon.
\end{align*}
Another approach: when $x \le y \le x + c$, you can write,
$$\int_{x}^{x + c}f(t) \; \mathrm{d}t - \int_{y}^{y + c}f(t) \; \mathrm{d}t \; \mathrm{d}t = \int_x^y f(t) \; \mathrm{d}t + \int_x^y f(t + c) \; \mathrm{d}t.$$
If you let $M$ be the maximum of $f$ over $[x, x + 2c]$. Then
$$\left|\int_{x}^{x + c}f(t) \; \mathrm{d}t - \int_{y}^{y + c}f(t) \; \mathrm{d}t \right| \le 2M|x - y|.$$
A: Even though Theo Bendit has given a good answer to your question already, I believe that things can be simplified considerably. Most importantly, you don't need to assume that $f$ is continuous. It just needs to be bounded (and integrable, of course), i.e. $\exists M\in\mathbb{N}, \forall t: \, |f(t)| \leq M$. 
Without loss of generality, assume $x<y<x+c$. We can write
$$ \bigg|F(x)-F(y)\bigg|=\bigg| \int_x^{x+c}f(t)dt -  \int_y^{y+c}f(t)dt\bigg|= \bigg| \int_x^yf(t)dt + \int_y^{x+c}f(t)dt -  \int_y^{y+c}f(t)dt \bigg|$$
$$\implies \bigg|F(x)-F(y)\bigg|= \bigg| \int_x^yf(t)dt  -  \int_{x+c}^{y+c}f(t)dt \bigg| \hspace{10px}(\star)$$
Using $(\star)$ and the triangle inequality, we have
$$\bigg| \int_x^yf(t)dt  -  \int_{x+c}^{y+c}f(t)dt \bigg| \leq \bigg| \int_x^yf(t)dt  \bigg|+\bigg|  \int_{x+c}^{y+c}f(t)dt \bigg| \leq \int_{x}^{y}\big|f(t)\big|dt+\int_{x+c}^{y+c}\big|f(t)\big|dt$$
Since $|f(t)| \leq M$, monotonicity of integration and the last inequality implies
$$\bigg| \int_x^yf(t)dt  -  \int_{x+c}^{y+c}f(t)dt \bigg| \leq |y-x|M + |y-x|M=2M|y-x| \hspace{10px} (\star\star)$$
So, you just need to take $\delta < \frac{\epsilon}{2M}$. Then $(\star)$ and $(\star\star)$ imply
$$|x - y| < \delta \implies |F(x) - F(y)| <\epsilon$$
Hence, $F(x)$ is uniformly continuous. $\fbox{Q.E.D.}$ 
If you assume $f$ to be locally bounded, as mentioned by Theo Bandit, the same idea shows that $F(x)$ will still be continuous, but not necessarily uniformly continuous.
