Closed subset of $\;C([0,1])$ 
$$\text{The set}\; A=\left\{x: \forall t\in[0,1] |x(t)|\leq \frac{t^2}{2}+1\right\}\;\;\text{is closed in}\;\, C\left([0,1]\right).$$

My proof: 
Let $\epsilon >0$ and let  $(x)_n\subset A$, $x_n\rightarrow x_0$. Then for $n\geq N$
 $\sup_{[0,1]}|x_n(t)-x_0(t)|\leq \epsilon$ for some $N$ and we have 
$|x_0(t)|\leq|x_0(t)-x_N(t)|+|x_N(t)|\leq\epsilon+\frac{t^2}{2}+1$
Then $|x_0(t)|\leq \frac{t^2}{2}+1$ for all $t\in[0,1]$
Is the proof correct?
Thank you. 

Yes, it is correct. An "alternative" proof would consist in writing $A$ as the intersection of the closed sets $F_t:=\{x,|x(t)|\leqslant \frac{t^2}2+1\}$. It also work if we replace $\frac{t^2}2+1$ by any continuous function. 
