A function $f:S \to \mathbb{R}$ defined on a convex subset $S$ of a real vector space is strictly-quasiconvex if for all $x, y \in S$ and $\lambda \in (0,1)$:
$$ f(\lambda x + (1 - \lambda)y) < \max\big\{f(x),f(y)\big\}. $$
A strictly quasiconvex function has a unique minimum in each convex subset $T\subseteq S$.
Proof: suppose by contradiction that $f$ has two minima, $x$ and $y$, in $T$. Define $z = x/2 + y/2$. Since $T$ is convex, $z\in T$ too. By definition of quaasiconvexity, either $f(z) < f(x)$ or $f(z) < f(y)$, so at least one of them is not a minimum.
Is the opposite also true, at least for continuous functions? I.e, is it true that if a continuous function $f$ has a unique minimum in every convex subset $T\subseteq S$, then it is strictly-quasiconvex in $S$?
I think I can prove a slightly weaker statement:
Suppose $f$ is not quasiconvex. There exist $x,y\in S$ and $\lambda \in (0,1)$ such that
$$ f(\lambda x + (1-\lambda)y) > \max\{ f(x),f(y) \}. $$
Assume that $f(x) \ge f(y)$, and define, for each $t\in[0,1]$,
$$ g(t)=f(tx+(1-t)y). $$
Since $g$ is continuous, there exists $t_x < \lambda$ such that $g(t_x)=f(x)$. Let $t_x$ be the largest such number less than $\lambda$. Similarly, let $s_x$ be the smallest number greater than $\lambda$ such that $g(s_x)=f(x)$. Formally,
\begin{align} t_x &= \sup\, [0,\lambda]\cap\{ t\in[0,1]:g(t) \le f(x) \}, \\ s_x &= \inf\, [\lambda,1]\cap\{ t\in[0,1]:g(t) \le f(x) \}. \end{align}
Now let $$T=\{z\in S: z = tx+(1-t)y,t\in[t_x,s_x] \}.$$ By construction, $T$ is convex. However, $f$ attains its minimum on $T$ at $s_xx+(1-s_x)y$ and $t_x x + (1-t_x)y$. (Note that $g(t)>f(x)=g(s_x)=g(t_x)$ for $t\in(t_x,s_x)$.) Thus, $f$ does not have a unique minimum on $T$.
If this argument is correct, it is possible to extend this to the result that you want; I believe it just boils down to breaking the analysis up into a few cases.