Why does the median minimize $E(|X-c|)$? Suppose $X$ is a real-valued random variable and let $P_X$ denote the distribution of $X$. Then
$$
E(|X-c|) = \int_\mathbb{R} |x-c| dP_X(x).
$$
The medians of $X$ are defined as any number $m \in \mathbb{R}$ such that $P(X \leq m) \geq \frac{1}{2}$ and $P(X \geq m) \geq \frac{1}{2}$.
Why do the medians solve
$$
\min_{c \in \mathbb{R}} E(|X-c|) \, ?
$$
 A: For every real valued random variable $X$,
$$
\mathrm E(|X-c|)=\int_{-\infty}^c\mathrm P(X\leqslant t)\,\mathrm dt+\int_c^{+\infty}\mathrm P(X\geqslant t)\,\mathrm dt
$$ 
hence the function $u:c\mapsto \mathrm E(|X-c|)$ is differentiable almost everywhere and, where $u'(c)$ exists, $u'(c)=\mathrm P(X\leqslant c)-\mathrm P(X\geqslant c)$. Hence $u'(c)\leqslant0$ if $c$ is smaller than every median, $u'(c)=0$ if $c$ is a median, and $u'(c)\geqslant0$ if $c$ is greater than every median.
The formula for $\mathrm E(|X-c|)$ is the integrated version of the relations  $$(x-y)^+=\int_y^{+\infty}[t\leqslant x]\,\mathrm dt$$ and $|x-c|=((-x)-(-c))^++(x-c)^+$, which yield, for every $x$ and $c$,
$$
|x-c|=\int_{-\infty}^c[x\leqslant t]\,\mathrm dt+\int_c^{+\infty}[x\geqslant t]\,\mathrm dt
$$
A: The following intends to complement Did's answer.

Claim
Denote by $M$ be the set of $X$'s medians. Then
  
  
*
  
*$M = [m_1, m_2]$ for some $m_1, m_2 \in \mathbb{R}$, such that $m_1 \leq m_2$.
  
*For every $m \in M$ and for every $x \in \mathbb{R}$ we have
  $$
 E\left(|X-m|\right) \leq E\left(|X-x|\right).
 $$
  (In particular, $m\mapsto E\left(|X-m|\right)$ is constant on $M$.)

Part 2's proof builds on Did's answer.
Proof


*

*It is known that $M \neq \emptyset$. Define
$$
 \begin{align}
 M_1 &:= \left\{t\in\mathbb{R}\ |\!:\ F_X(t) \geq \frac{1}{2}\right\}, \\
 M_2 &:= \left\{t\in\mathbb{R}\ |\!:\ P(X<t) \leq \frac{1}{2}\right\}.
 \end{align}
 $$
Then $M = M_1 \cap M_2$. It therefore suffices to show that $M_1 = [m_1, \infty)$ and that $M_2 = (-\infty, m_2]$, for some $m_1, m_2 \in \mathbb{R}$.
Since $\lim_{t\rightarrow-\infty}F_X(t) = 0$, $M_1$ is bounded from below. Since $\lim_{t\rightarrow\infty}F_X(t) = 1$, $M_1$ is an interval that extends to infinity. Hence $M_1 = (m_1,\infty)$ or $M_1 = [m_1,\infty)$, for some $m_1 \in \mathbb{R}$. It follows from $F_X$'s right-continuity that $m_1 \in M_1$. An analogous argument shows that $M_2 = (-\infty,m_2]$ (just verify that $t\mapsto P(X<t)$ is left-continuous).

*Define a function $f:\mathbb{R}\rightarrow\mathbb{R}$ as follows. For every $c \in \mathbb{R}$, set
$$
f(c) := E\left(|X-c|\right).
$$
We will begin by showing that $f$ is convex. Let $a, b \in \mathbb{R}$, and let $t \in (0,1)$. Then
$$
 \begin{align}
 f\left(ta+(1-t)b\right) &= E\left(\left|X-\left(ta+(1-t)b\right)\right|\right) \\
 &= E\left(\left|\left(tX-ta\right)+\left((1-t)X-(1-t)b\right)\right|\right) \\
 &\leq E\left(\left|\left(tX-ta\right)\right|+\left|\left((1-t)X-(1-t)b\right)\right|\right) \\
 &=E\left(\left|\left(tX-ta\right)\right|\right)+E\left(\left|\left((1-t)X-(1-t)b\right)\right|\right) \\
 &= t\ E\left(|X-a|\right) + (1-t)\ E\left(|X-b|\right) \\
 &= t\ f(a) + (1-t)\ f(b).
 \end{align}
 $$
Since $f$ is convex, then, by Theorem 7.40 of [1] (p. 157), there exists a set $A \subseteq \mathbb{R}$ such that $\mathbb{R}\setminus A$ is countable, and such that $f$ is finitely differentiable on $A$. Moreover, letting $m \in M$, and letting $x \in (-\infty, m_1)$, Theorem 7.43 of [1] (p. 158) yields that $f'$ is Lebesgue-integrable on $[x,m] \cap A$, and that
$$
 f(m) - f(x) = \int_{[x,m]\cap A} f'\ d\lambda.
 $$
Applying Did's answer, we find that $f'\leq 0$ on $[x,m]\cap A$. Hence $f(m) \leq f(x)$. Similar considerations show that, for every $x \in (m_2,\infty)$, $f(m) \leq f(x)$, and also that $f(m) = f(m_1)$ (implying that $f$ is constant on $M$, since $m$ was chosen arbitrarily in $M$).
(The argument of the last paragraph was suggested to me by copper.hat in their answer to a related question of mine.)
Q.E.D.

References
[1] Richard L. Wheeden and Antoni Zygmund. Measure and Integral: An Introduction to Real Analysis. 2nd Ed. 2015. CRC Press. ISBN: 978-1-4987-0290-4.
A: Let $f$ be the pdf and let $J(c) = E(|X-c|)$. We want to maximize $J(c)$. Note that $E(|X-c|) = \int_{\mathbb{R}} |x-c| f(x) dx = \int_{-\infty}^{c} (c-x) f(x) dx  + \int_c^{\infty} (x-c) f(x) dx.$
To find the maximum, set $\frac{dJ}{dc} = 0$. Hence, we get that,
$$\begin{align}
\frac{dJ}{dc} & = (c-x)f(x) | _{x=c} + \int_{-\infty}^{c} f(x) dx + (x-c)f(x) | _{x=c} - \int_c^{\infty} f(x) dx\\
& = \int_{-\infty}^{c} f(x) dx - \int_c^{\infty} f(x) dx = 0
\end{align}
$$
Hence, we get that $c$ is such that $$\int_{-\infty}^{c} f(x) dx = \int_c^{\infty} f(x) dx$$ i.e. $$P(X \leq c) = P(X > c).$$
However, we also know that $P(X \leq c) + P(X > c) = 1$. Hence, we get that $$P(X \leq c) = P(X > c) = \frac12.$$
EDIT
When $X$ doesn't have a density, all you need to do is to make use of integration by parts. We get that $$\displaystyle \int_{-\infty}^{c} (c-x) dP(x) = \lim_{y \rightarrow -\infty} (c-y) P(y) + \displaystyle \int_{c}^{\infty} P(x) dx.$$ Similarly, we also get that $$\displaystyle \int_{c}^{\infty} (x-c) dP(x) = \lim_{y \rightarrow \infty} (y-c) P(y) - \displaystyle \int_{c}^{\infty} P(x) dx.$$
A: Let $m$ be any median of $X$. Wlog, we can take $m=0$ (consider $X':=X-m$). The aim is to show $E|X-c|\ge E|X|$.
Consider the case $c\ge 0$. It is straightforward to check that $|X-c|-|X|=c$ when $X\le0$, and $|X-c|-|X|\ge -c$ when $X>0$. It follows that
$$
(|X-c|-|X|)\,I(X\le0)=c\,I(X\le0)\tag1
$$
and
$$(|X-c|-|X|)\,I(X>0)\ge-c\,I(X>0).\tag2
$$
Adding (1) and (2) and taking expectation yields
$$
E(|X-c|-|X|)\ge c\left[P(X\le0)-P(X>0)\right].\tag3
$$
The RHS of (3) equals $c\,[2P(X\le0)-1]$, which is non-negative since $c\ge0$ and zero is a median of $X$. The case $c\le0$ is reduced to the previous one by considering $X':=-X$ and $c':=-c$.
A: Let $Y=\left|X-c\right|$,
Then,
$$E(Y) = \int_0^\infty \left(1-F_Y(y)\right) dy$$
Note that, $F_Y(y) = F_X(c+y) - F_X(c-y),$
Thus
$$ \begin{align} E(Y) &= \int_0^\infty \big( 1-F_X(c+y) + F_X(c-y) \big) dy \\ \frac{d E(y)} {dc} &= \int_0^\infty \big(-f_X(c+y) + f_X(c-y) \big) dy \\ &=\int_0^\infty f_X(c-y) dy - \int_0^\infty f_X(c+y) dy 
\\ &= \int_{-\infty}^c f_X(x) dx - \int^{\infty}_c f_X(x) dx \\ & = F_X(c) - (1 - F_X(c)) \end{align} $$
Equating it to zero, we have,
$$F_X(c) = \frac{1}{2}$$
Hence median is the minimiser of $E(|X-c|)$.
