The problem:
Let $f: \mathbb{R}^n \rightarrow \mathbb{R}$ be a continuous function, $a \in \mathbb{R}^n$ be its local minimizer and $a$ is the only local minimizer in a neighborhood of itself. Prove that there is a neighborhood of $a$ satisfying that any sequence $(c_n)$ in that neighborhood converges to $a$ if the sequence $(f(c_n))$ converges to $f(a)$.
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I've tried to prove this proposition by contradiction. Due to continuity, it's easy to prove that if the sequence $(f(c_n))$ converges to $f(a)$ then there is no subsequence of $(c_n)$ converging to $b \neq a$. But I'm stuck after this.
This isn't true; if $f$ is constant $k$, then any sequence $(c_n)$ has the property that $f(c_n) \to k$, regardless of the behavior of the $c_n$'s.
If we assume instead that $f$ has a strict minimum, argue by contradiction. Suppose $c_n \not\to a$. This means there is a subsequence $c_{n_k}$ bounded away from $a$. Since all this is happening inside a bounded neighborhood, we can assume $c_{n_k}$ converges, by assumption to some $b \neq a$. But then $f(b) = f(a)$, contradicting strict minimality.