We consider in $\mathbb{R}^2$ the set of points $$\{M_1(-1,1),M_2(0,1), M_3(2,1),M_4(-1,0),M_5(1,0),M_6(2,0)\}$$
Let $\Omega$ a rectangular structure consisting of the heads $\{M_4(-1,0),M_6(2,0), M_3(2,1),M_1(-1,1)\}$
Let the two polygons $\sum_1 = \{M_4,M_5,M_2,M_1\}$ and $\sum_2 = \{M_5,M_6,M_3,M_2\}$
let the mesh $\{(e_1,\sum_1,Q_1),(e_2,\sum_2,Q_2)\}$ such $$Q_1=\{f : \mathbb{R}^2 \rightarrow \mathbb{R} , f(x,y) = a + bx + cy + d xy , a , b, c , d \in \mathbb{R}\}$$
1- Prove that $(e_1,\sum_1,Q_1)$ and $(e_2,\sum_2,Q_1)$ are two finite elements of Lagrange.
2- Let $V$ the finite-dimensional space that corresponds to the mesh. Prove that $V \nsubseteqq H^1(\Omega)$
My problem is to prove 2.
For any piecewise polynomial $v\in V$, $v$ restricted in each polygon has the form of $Q_1$: $$ v = \begin{cases} a_1+b_1 x+ c_1y + d_1 x y \quad\text{ in } \Sigma_1 \\ a_2+b_2 x+ c_2y + d_2 x y \quad\text{ in } \Sigma_2 \end{cases} $$
Now we wanna argue that there exits such $v\notin H^1(\Omega)$. For $v$ is definitely continuous in both $\Sigma_1$ and $\Sigma_2$ respectively, and it has $H^1$-regularity in both polygons. You can check $v$ restricted on $\Sigma_1$ is $L^2$ and so is its gradient, the same applies to $v$ restricted on $\Sigma_2$.
The only possible problematic part is where these two polygons intersect with each other: $\partial \Sigma_1 \cap\partial \Sigma_2 = M_2 M_5$.
The construction is pretty straight forward, we set $v = 1$ at node $M_2$, and zero on all other nodes, see if we can get a function that is continuous across the interface $M_2 M_5$ between these two polygons.
In $\Sigma_1$, we have: $$ \begin{cases} &a_1+&b_1 (-1)+ &c_1 1 + &d_1 (-1) &= v(M_1) = 0 \\ &a_1+ & &c_1 1 & &= v(M_2) = 1 \\ &a_1+ &b_1 1 & & &= v(M_5) = 0 \\ &a_1+ &b_1 (-1) & & &= v(M_4) = 0 \end{cases} $$ solving this leads to: $$ v\big|_{\Sigma_1} = y+xy. $$
In $\Sigma_2$, we have: $$ \begin{cases} &a_2+&b_2 2+ &c_2 1 + &d_1 2 &= v(M_3) = 0 \\ &a_2+ & &c_2 1 & &= v(M_2) = 1 \\ &a_2+ &b_2 1 & & &= v(M_5) = 0 \\ &a_2+ &b_2 2 & & &= v(M_6) = 0 \end{cases} $$ solving this leads to: $$ v\big|_{\Sigma_2} = y-\frac{xy}{2}. $$
Therefore, for a $v\in V$ satisfying that $v(M_2) = 1$ and zero at other nodes:
$$ v = \begin{cases} y+xy \quad\text{ in } \Sigma_1 \\ y-xy/2 \quad\text{ in } \Sigma_2 \end{cases} $$
Now it is up to you to check this function is not in $H^1$.
Update:
I already gave you the reference in the comments. You are not satisfied with its presentation, yet I think it is crystal clear: Ciarlet's The Finite Element Method for Elliptic Problems, please refer to the proof of Theorem 2.1.1 on pg40.
The result is commonly used for finite elements w/o citing any references, and it is very straightforward to prove. It can be formulated as follows:
Lemma: Suppose $\Sigma_1$ and $\Sigma_2$ are Lipschitz polygons, and $\partial\Sigma_1 \cap \partial \Sigma_2 = F$. Let $v_1\in H^1(\Sigma_1)$, $v_2\in H^1(\Sigma_2)$, and $v \in L^2(\Sigma_1\cup\Sigma_2\cup F)$ so that $v = v_1$ in $\Sigma_1$ and $v=v_2$ in $\Sigma_2$. Then if $v_1 = v_2$ on $F$, we have $v\in H^1(\Sigma_1\cup\Sigma_2\cup F)$.
Proof: Let $\Omega = \Sigma_1\cup\Sigma_2\cup F$. It suffices to show that the weak gradient of $v$ on the joint domain coincides with $\nabla v_1$ on $\Sigma_1$ and $\nabla v_2$ on $\Sigma_2$. Let $\phi \in C^{\infty}_c(\Omega)$, $$ \int_{\Omega} \frac{\partial v}{\partial x_i} \,\phi := - \int_{\Omega} v \,\frac{\partial \phi}{\partial x_i} . $$ This is the definition. The right side now can be divided into two parts: $$ - \int_{\Omega} v \,\frac{\partial \phi}{\partial x_i} = - \int_{\Sigma_1} v_1 \,\frac{\partial \phi}{\partial x_i} - \int_{\Sigma_2} v_2 \,\frac{\partial \phi}{\partial x_i} \\ = \int_{\Sigma_1} \frac{\partial v_1}{\partial x_i} \,\phi - \color{blue}{\int_{\partial \Sigma_1} v_1 \phi \,(n_{\partial \Sigma_1})_i\,dS}+ \int_{\Sigma_2} \frac{\partial v_2}{\partial x_i} \,\phi - \color{blue}{\int_{\partial \Sigma_2} v_2 \phi \,(n_{\partial \Sigma_2})_i\, dS}, $$ $n_{\partial \Sigma_k}$ is the unit outer normal vector to $\Sigma_k$ ($k=1,2$). Blue terms will cancel because $n_{\partial \Sigma_1} = -n_{\partial \Sigma_2}$ pointwisely, and $v_1 = v_2$ pointwisely as limit from their defined domain respectively (or use a trace argument if you like). Hence $$ \int_{\Omega} \frac{\partial v}{\partial x_i} \,\phi = \int_{\Sigma_1}\frac{\partial v_1}{\partial x_i} \,\phi + \int_{\Sigma_2} \frac{\partial v_2}{\partial x_i} \,\phi. $$ This tells us $\nabla v = \nabla v_1$ on $\Sigma_1$ and $\nabla v = \nabla v_2$ on $\Sigma_2$. Hence $$ |\nabla v|_{H^1(\Omega)}^2 = \|\nabla v\|_{L^2(\Omega)}^2 = \|\nabla v_1\|_{L^2(\Sigma_1)}^2 + \|\nabla v_2\|_{L^2(\Sigma_2)}^2 < \infty, $$ and the lemma follows.