Proving that a strongly convex function is coercive I am having trouble with this proof. I am given the following 2 definitions:

1) A function $f$ is coercive if $\lim_{||x|| \rightarrow \infty} f(x) = \infty$
2) A $C^2$ function $f$ is strongly convex if there exists a constant $c_0 > 0$ such that:  $(x - y)^T (\nabla f(x) - \nabla f(y)) \geq c_0 ||x - y||^2 \hspace{5mm}$ $\forall x,y \in \mathbb{R}^n$

The question is to show that if $f$ is strongly convex then it is coercive. 
I am only allowed to use basic theorems such as Taylor expansion, triangle inequality etc. However, I can use the fact that a function is coercive $\iff$ all its level sets are compact.
My instinct is that the answer can be obtained by a doing a Taylor expansion and manipulating the result, but I've been stuck for days using this approach. Any help would be greatly appreciated. 
 A: Lemma 1: Let $f:\mathbb R^n\to \mathbb R$ be a Fréchet-differentiable function. $f$ is convex iff $\forall x,y\in \mathbb R^n, (\nabla f(y)-\nabla f(x))^T(y-x)\geq 0$.
Lemma 2: Let $f:\mathbb R^n\to \mathbb R$ be a Fréchet-differentiable function. $f$ is convex iff $\forall x,y\in \mathbb R^n, f(y)\geq f(x)+\nabla f(x)^T(y-x)$

Let us prove first that your definition of "strongly convex" implies convexity of $\displaystyle g:x\mapsto f(x)-\frac{c_0}2\|x\|^2$
Indeed, $\begin{aligned}[t](\nabla g(y)-\nabla g(x))^T(y-x)&=(\nabla f(y)-c_0y - (\nabla f(x)-c_0x))^T(y-x)\\
&=(\nabla f(y)-\nabla f(x))^T(y-x) - (c_0y-c_0x)^T(y-x) \\
&= (\nabla f(y)-\nabla f(x))^T(y-x)-c_0\|y-x\|^2 \\
&\geq 0 \quad \text{ with your definition}
\end{aligned}$
By Lemma 1, $g$ is convex.

By Lemma 2, $$\begin{aligned}[t]
g(y)&\geq g(x)+\nabla g(x)^T(y-x) \\
f(y)-\frac{c_0}2\|y\|^2 &\geq f(x)-\frac{c_0}2\|x\|^2 + (\nabla f(x)-c_0x)^T(y-x)
\end{aligned}$$
Let $\alpha =\nabla f(x)-c_0x$. Then 
$$f(y)\geq \frac{c_0}2\|y\|^2+ \alpha^Ty+K_x$$ where $K_x$ is a constant that depends only on $x$.
By Cauchy-Schwarz, $\alpha^Ty \geq -\|\alpha\|\|y\|$, hence
$$f(y)\geq \frac{c_0}2\|y\|^2 -\|\alpha\|\|y\|+K_x$$
The right-hand side goes to $\infty$ as $\|y\|\to \infty$ and we're done.
