There's a part of proof in a theorem, which I don't understand. I'll write down the theorem and the proof until the point, where I'm stuck
Theorem: There exists a constant $c=c(Q,k,p,n)$ such that $$|\tilde{v}|_{k+1,p,Q}\leq\|\tilde{v}\|_{k+1,p,Q}\leq c |\tilde{v}|_{k+1,p,Q}$$ for any $\tilde{v}\in W^{k+1,p}(Q)/P_k$.
Proof:
Let $\varphi_1,...,\varphi_N$ be a basis of $P_k$, $f_1,...,f_N$ the dual basis, so $f_i(\varphi_j)=\delta_{ij}$.
First we'll show, that there exists a constant $c=c(Q,k,p,n)$ such that $$\|v\|_{k+1,p,Q}\leq c (|v|_{k+1,p,Q}+\sum_{i=1}^N|f_i(v)|)\ \ \forall v\in W^{k+1,p}(Q)\ \ \ (*)$$ The proof will be done by contradiction, so we'll suppose that $(*)$ doesn't hold. That implies, that $$\forall m\in \mathbb{N}\ \exists v_m\in W^{k+1,p}(Q):\ \|v_m\|_{k+1,p,Q}> m (|v_m|_{k+1,p,Q}+\sum_{i=1}^N|f_i(v_m)|)$$ Without loss of generality: $\|v_m\|_{k+1,p,Q}=1$. This implies: $$\lim_{m\rightarrow\infty}\left(|v_m|_{k+1,p,Q}+\sum_{i=1}^N|f_i(v_m)|\right)=0$$
Now comes the part, which I don't understand:
The embedding of $W^{k+1,p}(Q)$ into $W^{k,p}(Q)$ is totally continuous. This implies: $\exists v\in W^{k,p}(Q)$, $\{v_m\}$ subsequence s.t. $$\|v_m-v\|_{k,p,Q}\rightarrow0$$
The textbook actually doesn't state "subsequence of what sequence" $\{v_m\}$ is. I suppose it's supposed to be the subsequence of $\{v_m\}$, named the same. Is my assumption correct? And why does a totally continuous embedding imply this fact?