Let $$H_0^1(0,1)=\{f\in W^{1,2}(0,1):f(0)=0\}$$ and a norm $$\| f\|=\left (\int_0^1 |f'(x)|^2\mathrm{d}x\right )^{1/2}$$
be given. I want to show that if a sequence $(u_n)_{n\in\mathbb{N}}$ in $(H_0^1(0,1),\|\cdot\|)$ converges to $u$ and furthermore $u\in W^{1,2}(0,1)$ then $u$ must be in $H_0^1(0,1)$. In other words, $H_0^1(0,1)$ is a closed subspace of $W^{1,2}(0,1)$.
I tried the following:
The mean value theorem for sobolev functions is analogue to the one in the space of continuous functions on compact intervalls $$f(x)=f(0)+\int_0^xf'(y)\,\mathrm{d}y$$
Now we have by assumption that for every $\epsilon >0$ there is an $N\in\mathbb{N}$ such that for all $n\ge N$ $$\|f-f_n\|^2=\int_0^1((f-f_n)')^2\mathrm{d}x=\int_0^1(f'-f_n')^2\mathrm{d}x<\epsilon$$
Now using the MVT for sobolev spaces I can obtain $$|f(x)-f_n(x)|-|f(0)-f_n(0)|=\int_0^x (f'-f'_n)\mathrm{d}y<\epsilon$$
Now since, $f_n(0)=0$ and $|f(x)-f_n(x)|$ converges to zero (we assumed that $f$ would be in the sovolev space and therefore also converges by supremum norm), I can conclude that $f(0)=0$.
I have no idea if this is correct and appreciate any corrections/improvement.
Here is an outline of what you can do. The details aren't difficult to fill in but would make this answer too long.
First, let $W = W^{1,2}(0,1)$ and let $H = H_0^1(0,1)$. Define norms $$ \|f\|_W = \left( \int_0^1 |f|^2 + |f'|^2 \, dx \right)^{1/2} \quad \text{and} \quad \|f\|_H = \left( \int_0^1 |f'|^2 \, dx\right)^{1/2}.$$ Functions in $W$ are (absolutely) continuous, so identify each function with its continuous representative. Then the fundamental theorem of calculus gives you $$ \tag{1}\sup_{x \in [0,1]} |f(x)| \le \sqrt{2}\|f\|_W$$ and all $f \in W$ and $$ \sup_{x \in [0,1]} |f(x)| \le \|f\|_H$$ for all $f \in H$. Moreover, after integrating the second inequality you get $$ \tag{2} \|f\|_W \le \sqrt 2 \|f\|_H$$ for all $f \in H$. I'm not sure if the constant $\sqrt 2$ is best--it's not important for the argument.
Let $\{f_n\} \subset H$ be a Cauchy sequence in the norm $\| \cdot\|_H$. Then $(2)$ tells you $\{f_n\}$ is also Cauchy in the norm $\|\cdot \|_W$. Since $W$ is complete there exists $f \in W$ with $\|f_n - f\|_W \to 0$.
Now refer back to $(1)$: since $$ |f(0)| = |f_n(0) - f(0)| \le \sup_{x \in [0,1]} |f_n(x) - f(x)| \le \sqrt{2} \|f_n - f\|_W \to 0$$ you get $f(0) = 0$ so that $f \in H$. In light of the obvious inequality $\|\cdot\|_H \le \|\cdot\|_W$ you have $\|f_n - f\|_H \to 0$ too, implying that $H$ is complete with respect to its norm.
Since the norm of $H$ is induced by the inner product $$\langle f,g \rangle = \int_0^1 f'(x) g'(x) \, dx$$ you conclude $H$ is a Hilbert space.