Thanks to those that have already responded, you were very helpful. Here I will give the solution I have come up with, with more exposition than is provided in some of the other responses.
First we need the following lemma:
Lemma: $\lim_{\|x\| \to \infty} f(x) = \infty$. Some authors refer to this as $f$ being coercive.
Proof: Let $x_0 \in \mathbb{R}^n$ and let $v$ be a subgradient of $g$ at $x_0$, i.e. $x_0 \in \partial g(x_0)$. By equivalence of norms in finite-dimensional vector spaces, there exists a constant $c > 0$ such that $\|x\|_2 \leq c \|x\|$ for all $x \in \mathbb{R}^n$. By Cauchy-Schwarz and the trinagle inequality, for $\|x\| > 0$ we have
$$
\begin{align*}
\frac{| v^T(x - x_0) |}{\frac{m}{2}\|x\|^2}
&\leq \frac{\|v\|_2 \|x - x_0\|_2}{\frac{m}{2}\|x\|^2} \\
&\leq \frac{\|v\|_2 \|x\|_2 + \|v\|_2 \|x_0\|_2}{\frac{m}{2}\|x\|^2} \\
&\leq \frac{c\|v\|_2 \|x\| + \|v\|_2 \|x_0\|_2}{\frac{m}{2}\|x\|^2} \\
&= \frac{2c\|v\|_2 \|x\|}{m} \frac{1}{\|x\|} + \frac{2\|v\|_2 \|x_0\|_2}{m} \frac{1}{\|x\|^2}
\end{align*}
$$
The far right hand side of this inequality $\to 0$ as $\|x\| \to \infty$, which implies that $v^T(x - x_0) + \frac{m}{2} \|x\|^2 \to \infty$ as $\|x\| \to \infty$. Now we use the definition of subgradient:
$$
\begin{align*}
v^T(x - x_0) &\leq g(x) - g(x_0) \\
v^T(x - x_0) + \frac{m}{2}\|x\|^2 &\leq g(x) + \frac{m}{2}\|x\|^2 - g(x_0) \\
v^T(x - x_0) + \frac{m}{2}\|x\|^2 + g(x_0) &\leq f(x)
\end{align*}
$$
The left hand side of this $\to \infty$ as $\|x\| \to \infty$, so we conclude that $f(x) \to \infty$ as $\|x\| \to \infty$. $\square$
On to the main result. First, assume that $A$ is unbounded. If it is bounded, then it is compact, and the result follows immediately. There are 2 mutually exclusive possibilities:
Case 1: $f$ has a minimizer on $A$, in which case it is unique (see this thread).
Case 2: $f$ does not have a minimizer on $A$.
Assume we have case 2. Let $f^\star := \inf_{x \in A} f(x)$. $f^\star < \infty$ by assumption. Let $(x_k)$ be a sequence in $A$ such that $f(x_k) \to f^\star$. We now have two mutually exclusive subcases:
Subcase 2.1: $\sup_k \|x_k\| = d < \infty$. Define $B_d := \{ x \in \mathbb{R}^n \ : \ \|x\| \leq d\}$. Then for all $k$, $x_k \in \{ A \cap B_d \}$ which is closed and bounded and hence compact. Therefore $(x_k)$ converges in $A$, i.e. $x_k \to x^\star$ for some $x^\star \in A$. Continuity of $f$ then implies $f(x^\star) = f^\star$, which is a contradiction.
Subcase 2.2: $\sup_k \|x_k\| = \infty$. This implies $\|x_k\| \to \infty$, and by the Lemma this implies $f(x_k) \to \infty$ which implies $f^\star = \infty$ which contradicts $f^\star < \infty$.
Thus we conclude that Case 2 cannot occur, and therefore Case 1 must occur.
EDIT: After writing all of this out, it is clear that $f$ strongly convex is a stronger assumption than we require. $f$ strictly convex and coercive is sufficient for $f$ to have a unique global minimum on the convex set $A$.
Best Answer
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.