How does the inner product (-,-) with norm ||.|| relate to bilinear and linear forms in the FEM?

I am attempting to prove the Lax-Milgram lemma for the weak formulation of the finite element method. However I first need to prove continuity of the bilinear and linear forms ($a(u,v)$ and $l(v)$), plus coercivity of the bilinear form. I am struggling to understand how they connect to the norm. I know the $u$ and $v$ lie in the hilbert space $H^1(\Omega)$ with the $L^2$ norm.

I have that $a(u,v)=\int_{\Omega}\bigtriangledown u\cdot\bigtriangledown v$

so does that make the norm associated with it $||\bigtriangledown u\cdot\bigtriangledown v||$?

If you wish to use Lax-Milgram lemma, you must show continuity of the bilinear form $a$ in the space which is relevant to the posed problem. Suppose that you are solving $$- \Delta u = f$$ in domain $\Omega$ with the boundary condition $u|_{\partial \Omega}=0$. Then the weak formulation reads: find $u \in H^1_0(\Omega)$ such that $$(\nabla u, \nabla v) = (f,v)$$ holds for every $v \in H^1_0(\Omega)$. Now $H^1_0(\Omega)$ is a Hilbert space with the norm $$\|w\|_{H^1_0(\Omega)}:=\|\nabla w\|_{L^2(\Omega)}$$ and, therefore, the continuity follows immediately from the Cauchy-Schwarz inequality: $$(\nabla u, \nabla v) \leq \|\nabla u\|_{L^2(\Omega)} \|\nabla v\|_{L^2(\Omega)}.$$ Moreover, coercivity is obvious: $$(\nabla u, \nabla u) = \|\nabla u\|_{L^2(\Omega)}^2.$$
Another example. Suppose that you are solving a pure Neumann problem: $$- \Delta u = f$$ in domain $\Omega$ with the boundary condition $(\nabla u \cdot n)|_{\partial \Omega}=0$. Then the weak formulation is: find $u \in H^1(\Omega)$ such that $$(\nabla u, \nabla v) = (f,v)$$ holds for every $v \in H^1(\Omega)$. Now the respective norm is $$\|w\|_{H^1(\Omega)}^2 := \|\nabla w\|_{L^2(\Omega)}^2 +\|w\|_{L^2(\Omega)}^2$$ Continuity can be proven somewhat similarly as before but now you cannot show the coercivity: $$(\nabla u, \nabla u) = \|\nabla u\|_{L^2(\Omega)}^2 \geq~???$$ Thus, the solution to this problem is not unique. In order to make the solution unique, you can for example include the constraint $$\int_\Omega u \,\mathrm{d}x = 0.$$ With this constraint you can use the Poincaré-Friedrich's inequality to get $$(\nabla u, \nabla u) = \frac 12 \|\nabla u\|_{L^2(\Omega)}^2+\frac 12 \|\nabla u\|_{L^2(\Omega)}^2 \geq \frac 12 \|\nabla u\|_{L^2(\Omega)}^2 + \frac{1}{2C}\|u\|_{L^2(\Omega)}^2 \geq \frac{1}{2 \max(1,C)} \|u\|_{H^1(\Omega)}^2.$$