Why is the zeroth homology group in singular homology isomorphic to Z? Thanks to freakish's correction below:
Let $X$ be a path-connected topological space. Consider a chain complex $C_n$ and boundary maps $\delta_n: C_n \to C_{n-1}$. Then for $n=0$, we have $\delta_0: C_0 \to 0$, and thus $ker(\delta_0)=C_0$. Define $H_0(X)$ to be $C_0/im(\delta_1)$, I want to know why $H_0(X)$ is isomorphic to $\mathbb{Z}$?
 A: Throught I will assume that by "homology" you mean "singular homology".

Why is the zeroth homology group isomorphic to Z?

It is not. That depends on the space.

I know that $H_0(X)=Ker(X)/Im(X)$ and $Ker(X)$ is the entire space.

That... doesn't make any sense. We consider a chain complex $C_n$ and boundary maps $\delta_n:C_n\to C_{n-1}$. Then indeed for $n=0$, in the singular case $C_{-1}=0$ by definition, and thus $\ker\delta_0=C_0$. Or we ignore negative indexes and simply define $H_0(X)$ to be $C_0/\text{im }\delta_1$. I assume this is what you've meant.
So what exactly $H_0(X)$ is?
tl;dr; $H_0(X)$ is isomorphic to a direct sum $\bigoplus\mathbb{Z}$ with one $\mathbb{Z}$ for each path-component of $X$. In particular $H_0(X)\simeq\mathbb{Z}$ only for path-connected spaces.
Longer explanation: we need to understand what $\delta_1:C_1\to C_0$ is. Given a singular simplex $s:\Delta^1\to X$ we have that $\delta_1(s)=s_1-s_0$ where $s_1,s_0:\Delta^0\to X$ are constant maps at ends of $s$ simplex (it has ends, because $\Delta^1$ is an interval). Note that since $\Delta^1$ is path connected, then ends of $s$ lie in the same path component of $X$. So let $[x]$ represent a path component of $x\in X$ and $P(X)$ be the set of all path components of $X$. Consider $V$ to be the free abelian group generated by $P(X)$, which is isomorphic to $\bigoplus_{[x]\in P(X)}\mathbb{Z}$. Then consider
$$\tau:V\to C_0/\text{im }\delta_1$$
$$\tau([x])=s_x + \text{im }\delta_1$$
where on the right side we have the constant map $s_x:\Delta^0\to X$, constant at $x\in X$. This doesn't depend on the choice of $x$, because if $y$ is path connected to $x$ via $\lambda:\Delta^1\to X$, then
$$s_x+\text{im }\delta_1=s_y+s_x-s_y+\text{im }\delta_1=s_y+\text{im }\delta_1$$
the last equality because $s_x-s_y=\delta_1(\lambda)\in\text{im }\delta_1$ and so it vanishes in the quotient.
I leave as an exercise that $\tau$ is bijective. Therefore it is an isomorphism.
