We must show the set $S = \{x\in \mathbb{R}^n \:|\: f(x)\leq f(x_0) \}$ is bounded for some $x_0 \in \mathbb{R}^n$ where $f$ is strongly convex. That is, $||y||_2\leq C$ for some constant $C$ for all $y\in S$. When we say $f$ is strongly convex, this means the eigenvalues of the Hessian matrix $\nabla^2f$ are all positive and bounded as $0<m\leq \lambda \leq M$.
Because strongly convex functions can take on a variety of different shapes, there isn't a way to calculate an exact bound $C$. Instead, there are two main approachs. 1) Proof by contradiction and 2) Showing the set $S_0= \{x\in \mathbb{R}^n \:|\: f(x)=f(x_0)\}$ is bounded and also show that any vector with a length longer than this bound not in $S$.
From my understanding of a Hessian matrix given a point in $\mathbb{R}^n$, $(\nabla^2f)v$ is a vector that shows in what direction the gradient vector at this point is nudged as one travels in the direction $v$. If you travel in the direction of an eigenvector of $\nabla^2f$ then the gradient vector is nudged in the exact same direction. $f$ being strongly convex implies that at any point in the domain traveling in the direction of a vector $v$ will nudge the gradient vector in a direction similar to $v$ (same orthant).
I don't know where to go from here.
This result can be easily obtained through the following corollary: Given closed convex function $f$, if there exists one $\alpha_0$ such that non-empty level set $\{x \mid f(x) \leq \alpha_0\}$ is bounded, then every non-empty level set (i.e., $\{x \mid f(x) \leq \alpha\}$ for any $\alpha$) is also bounded.
This corollary is derived from the fact that all nonempty level sets of $f$ have the same recession cone.
Thus what we need to do is find a non-empty and bounded level set for function $f$. Obviously, due to the strongly convexity, the level set $\{x \mid f(x) \leq \min f(x)\}$ contains only one element and thus is bounded. Therefore, level set $\{x \mid f(x) \leq f(x_0)\}$ is bounded.
One can refer to Rackafaller's book ``Convex Analysis'' (Chapter 8) for more details.