If local minimum whenever assuming value then global

Question

Let $f:\mathbb{R}^n\to\mathbb{R}$ be continuous, $x_0\in\mathbb{R}^n$ and $y_0 = f(x_0)$. Suppose that whenever $f(x) = y_0$, $x$ is a local minimum of $f$. Prove that $x_0$ is a global minimum of $f$.
I figured I should be using the connectivity of the domain to justify that, but I can only think of proving it by contradiction, and it is getting me nowhere...

Answer 1

Hint

Suppose that it exists $a \in \mathbb R^n$ with $f(a) <y_0$ and consider the segment $[x_0,a]$.

Consider $T=\{t \in [0,1] \mid f((1-t)x_0 + ta) \ge y_0\}$. $T$ is not empty as $0 \in T$, bounded by $1$ and closed as $f$ is continuous. Therefore $T$ has an upper bound $b <1$ such that $f(x_b)= y_0$ where $x_b = (1-b)x_0+ba$. By hypothesis, $x_b$ is a local minimum of $f$. A contradiction as $f((1-(b+\epsilon))x_0 + (b + \epsilon)a) < y_0$ for $\epsilon >0$ small enough.