You made a small mistake in your computation that $\Phi$ is continuous in either of the arguments --- but you're on the right track! One starts as you did by computing
$$
|\Phi(p_n,q)-\Phi(p,q)|=\Big|\int_0^1(p_n(x)-p(x))q(x)\,dx\Big|\leq\int_0^1|p_n(x)-p(x)|\,|q(x)|\,dx\,.\tag{1}
$$
What we know is that $p_n\to p$ as $n\to\infty$, that is, $\lim_{n\to\infty}\int_0^1|p_n(x)-p(x)|\,dx=0$. This is almost the expression in (1) if it were not for $q$. Fortunately, as a polynomial $q$ is continuous meaning it is bounded on any bounded interval. This yields
\begin{align*}
|\Phi(p_n,q)-\Phi(p,q)|&\leq\int_0^1|p_n(x)-p(x)|\,|q(x)|\,dx\\
&\leq\int_0^1|p_n(x)-p(x)|\,\big(\max_{x'\in[0,1]}|q(x')|\big)\,dx\\
&=\underbrace{\big(\max_{x'\in[0,1]}|q(x')|\big)}_{<\infty}\;\underbrace{\int_0^1|p_n(x)-p(x)|\,dx}_{=\|p_n-p\|\to 0}\to 0
\end{align*}
as $n\to\infty$. One argues similarly for the second component. Side note: the key estimate here is a special case of Hölder's inequality which would also deal with the domain of $\Phi$ being equipped with any other $p$-norm.
For discontinuity of $\Phi$ the hint @blamethelag gave in their comment is a good starting point. Just plugging this in suggests that if we can find polynomials $(p_n)_{n\in\mathbb N}$, $p$ such that $\|p_n-p\|\to 0$ but
$$
\Phi(p_n,p_n)=\int_0^1|p_n(x)|^2\,dx\not\to\int_0^1|p(x)|^2=\Phi(p,p)
$$
then we are done. Also from the proof given above we know that any such counterexample has to satisfy $\max_{x'\in[0,1]}|p_n(x')-p(x')|\not\to 0$ as $n\to\infty$ (because else Hölder's inequality would show that $\Phi(p_n,p_n)$ converges to $\Phi(p,p)$). One of the standard counterexamples in this context is the sequence of polynomials $p_n(x):=x^n$ with limit $p(x):=0$. However, a quick computation shows this does not quite work yet, but maybe it does after tweaking things a bit.
Let us define $p_n(x):=a_nx^n$ where, for now, $(a_n)_{n\in\mathbb N}$ is an arbitrary sequence of numbers. By plugging this in we want to see what the conditions on $a_n$ are to make for a valid counterexample. In other words we need
\begin{align*}
0=\lim_{n\to\infty}\|p_n-p\|&=\lim_{n\to\infty}\int_0^1|a_n x^n-0|\,dx\\
&=\lim_{n\to\infty}|a_n|\int_0^1 x^n\,dx=\lim_{n\to\infty}|a_n|\Big[\frac1{x^{n+1}}\Big]_0^1=\lim_{n\to\infty} \frac{|a_n|}{n+1}
\end{align*}
but
\begin{align*}
0\neq\lim_{n\to\infty}|\Phi(p_n,p_n),\Phi(p,p)|&=\lim_{n\to\infty}\Big|\int_0^1|a_nx^n|^2\,dx\Big|\\
&=\lim_{n\to\infty}|a_n|^2\int_0^1x^{2n}\,dx=\lim_{n\to\infty}\frac{|a_n|^2}{2n+1}\,.
\end{align*}
Thus if we can find numbers which
grow slower than $n+1$ but not faster than $2n+1$ we are done. The simplest choice here is probably $a_n=\sqrt{n}$, that is, $p_n(x):=\sqrt nx^n$ converges to the zero function in $\|\cdot\|$, but $\Phi(p_n,p_n)$ does not converge to zero $=\Phi(p,p)$.
