Prove that $A= \left\lbrace f \in X: \int_0^1f(x) dx>1 \right\rbrace$ is an open set in $(X,d)$.

Note that the metric d is equivalent to the usual maximum metric on C[0,1] (because $e^{-x^2}$ attains non-zero min and max values on [0,1]), so for the purposes of your questions you may as well replace d by that metric. For Q1) consider a sequence of "tent functions" whose tent width tends to 0, but whose height remains constant. For Q2) just note that the complement of A is closed in the $p$ metric and use what you have already proven to conclude that then it is also closed in the $d$ metric.

Let $f_n$ be the function whose graph is given by joining the points $(0,0), ({1 \over n}, 1), ({2 \over n}, 0), (1,0)$. Let $f= 0$.

Then $f_n \to f$ with distance $p$, but $d(f_n,f) \to 1$.

Also, if we let $Lf = \int_0^1 f(x)dx$ then we have $|Lf-Lg| \le p(f,g) \le e \cdot d(f,g)$ and so $L$ is continuous with respect to both metrics. Hence $A=L^{-1}((1,\infty))$ is open.

Let $f_n(x)=x^n$ and $f(x)=0$. It is easy to see $$ \int_0^1|f_n(x)-f(x)|dx=\frac1{n+1}\to 0 $$ as $n\to\infty$. However $$ d(f_n,f)=\max_{x\in[0,1]}x^ne^{-x^2}=\left(\frac{n}{2e}\right)^{n/2}\to\infty $$ as $n\to\infty$.