Suppose we have a compact set $K \subset \mathbb{R}$ and a sequence of continuous functions $f_n: K \rightarrow \mathbb{R}$. Let $f$ be the uniform (and hence continuous) limit of $(f_n)_n$. Assume further that each $f_n$ has a unique maximum $max_n$ and $f$ has a unique maximum $max$.
The question is whether (and why/why not) the sequence of maxima $(max_n)_n$ converges to $max$.
I note $max_n=m_n$ and $max=m$.
a) Let $\varepsilon>0$. By the uniform convergence, there exists a $N$ such that if $n\geq N$, we have for all $x\in K$ $$f(x)-\varepsilon\leq f_n(x)\leq f(x)+\varepsilon$$
b) As $K$ is compact and $f_n$ continuous, there exists $u_n\in K$ such that $f_n(u_n)=m_n$. We have for $n\geq N$ $$ m_n=f_n(u_n)\leq f(u_n)+\varepsilon\leq m+\varepsilon $$
c) There exists $u\in K$ such that $f(u)=m$. We have $$m_n\geq f_n(u)\geq f(u)-\varepsilon=m-\varepsilon$$
d) We have proven that for any $\varepsilon>0$, there exists a $N$ such that for $n\geq N$, $|m_n-m|\leq \varepsilon$, and we are done.