From Brezis chapter 8. When he goes through some examples of BVP, for one particular example (inhomogeneous Dirichlet problem) he constructs a set, \begin{align} K:=\{v\in H^{1}(I)\,|\,v(0)=\alpha\quad\text{and}\quad v(1)=\beta\}, \end{align} where $I=(0,1)$. He states the set $K$ is closed and convex.
To show convexity, suppose $u,v\in K$. Then $tv+(1-t)u$ evaluated at $0$ and $1$ gives, \begin{align} tv(0)+(1-t)u(0)=\alpha,\\ tv(1)+(1-t)u(1)=\beta. \end{align} Hence, $tv+(1-t)u\in K$ and so $K$ is convex.
To show $K$ is closed is where I am a little stuck. Suppose $(v_{n})\in K$ is a convergent sequence with limit $v\in H^{1}(I)$. Then for all $\varepsilon>0$ there exists $n_{0}$ such that for $n>n_{0}$, $\|v_{n}-v\|_{H^{1}}<\varepsilon$. Now convergence in $H^{1}$ implies there exists a subsequence $(v_{n_{k}})$ which converges pointwise to $v$. Hence, since $v_{n_{k}}(0)=\alpha$ and $v_{n_{k}}(1)=\beta$ for all $k$ then $v(0)=\alpha$ and $v(1)=\beta$. Therefore $v\in K$ and so $K$ is closed.
Is this proof correct?
Define $$ Av = v(a),\;\;\; Bv=v(b). $$ These are continuous linear functionals on $H^1(I)$ because, for example, $$ v(a) = -\left.\frac{b-t}{b-a}v(t)\right|_{t=a}^{b}\\= -\int_{a}^{b}\frac{d}{dt}\left(\frac{b-t}{b-a}v(t)\right)dt \\ = \int_{a}^{b}\frac{1}{b-a}v(t)-\frac{b-t}{b-a}v'(t)dt, $$ and $$ |Av|=|v(a)| \le C(\|v\|_{L^2}+\|v'\|_{L^2}) $$ Similary, $|Bv|\le C\|v\|_{H^2}$ for a constant $C$. Therefore, $K=\{ v\in H^1 : A(v)=B(v)=0 \}$ is closed as it is the intersection of two closed sets, both being the inverse image of $\{0\}$ under continuous linear functionals.