real analysis – If $f_nto f$ uniformly, then $min_{[a,b]}f_n to min_{[a,b]}f$

For $f_n,f$ continuous on $(a,b)$, and $f_n to f$ uniformly, define $M_n=max_{(a,b)}f_n$ and $M=max_{(a,b)}f$, then it was asked to show that $M_n to M$. And it was also asked is it also true for sequence of minima. I did the first part as follows:

$f_n(x)-f(x)leq max_{(a,b)}lvert f_n(x)-f(x) rvert$,

$Rightarrow f_n(x)leq max_{(a,b)}lvert f_n(x)-f(x) rvert + f(x)$,

Then first taking maximum on right side and then on left side, we will get,

$M_n-Mleq max_{(a,b)}lvert f_n(x)-f(x) rvert$.

Similarly, we can also show that,

$M-M_nleq max_{(a,b)}lvert f_n(x)-f(x) rvert$, which gives,

$lvert M_n-M rvert leq max_{(a,b)}lvert f_n(x)-f(x) rvert$.

By uniform continuity, it is evident that right hand side of above equation goes to zero, which implies $M_n to M$.

But I am having difficulty in showing the second part of question. I guess it is true and I thought proof will be similar but the same idea doesn’t seem to work. Any hint. Thanks.