Show that in the metric space $\left(C([0,1]), d_{\infty}\right)$, the subset $$ A=\{f \in C([0,1]): f(t)>t, \forall t \in[0,1]\} $$ is open.
My try: I take $x\in A$ and $\epsilon=f(t)-t$.
I take $g$ in the ball $B_{\epsilon}(f)$.
So $f(t)-g(t)<\max|f(t)-g(t)|< \epsilon$ for all $t$.
So $f(t)-g(t)< f(t)-t$, and hence $g(t)> t$.
Is it right?
On
The problem with your proof is that it is not clear that a $t$-independent positive $\epsilon$ exists, such that $f(t)-t>\epsilon$, especially because you have not used the fact that the set of continuous functions on a compact set ($[0,1]$) is under question. In the following, we prove that such $\epsilon>0$ exists.
We prove that for any $h\in A$, $\inf_{0\le x\le 1}\{h(x)-x\}>0$. Assume by contrary that $\inf_{0\le x\le 1}\{h(x)-x\}=0$. Then there exists a sequence $\{x_n\}_{n=1}^\infty\in[0,1]^\infty$ such that $\lim_{n\to\infty} h(x_n)-x_n=0$. Since $[0,1]$ is compact, there exists some subsequence $\{x_{y_n}\}_{n=1}^\infty\in[0,1]^\infty$ that converges to some $x^*\in[0,1]$. This, by definition of continuity of $h(x)$, implies that $h(x^*)=x^*$, which is a contradiction. Hence, $\inf_{0\le x\le 1}\{h(x)-x\}>0$ and the proof is complete. Now, simply regard any $\epsilon<\inf_{0\le x\le 1}\{h(x)-x\}$ and finish your proof cheerfully.
Shedding light on the problem with your proof
If the set of continuous functions is defined over any non-compact set $S$, it is possible to find functions $h\in A$ such that $\inf_{S}\{h(x)-x\}=0$. For example, consider the set of continuous functions over $(0,1)$. Then, for $h(t)=\sqrt t$ we have $$\inf_{t\in(0,1)}h(t)-t=\inf_{t\in(0,1)}\sqrt t-t=0,$$implying that no $\epsilon>0$ exists such that $h(t)-t>\epsilon$.
Hint
If $f\in A$, then $g(t):=f(t)-t$ is continuous and $g(t)>0$ for all $t\in [0,1]$. There should be some reason, using that $g$ is continuous on a closed interval, why there exists some $\epsilon>0$, such that $g(t)\geq\epsilon>0$ for all $t\in [0,1]$.