Let $V$ and $K$ be Banach spaces (with norms $\|\cdot\|_V$ and $\|\cdot\|_K$ resp.) and suppose that there is a compact linear embedding $K\hookrightarrow V$. Furthermore, let $P_n$ be a family of projections on $V$ whose rank is finite and that for all $u\in V$, $\lim_{n\rightarrow\infty} \|P_n u-u\|_V=0$.
Is there a bounded sequence $s_n$ that tends to zero and \begin{equation} \|P_n u-u\|_V \leq s_n\|u\|_K \end{equation} for all $u\in K$?
The motivation to this question is to generalize the usual finite element interpolation estimate "$\|I_h u -u\|_1 \leq C h^{k-1}|u|_k$", where $k$ is the order polynomial order and $h$ is the mesh size of the FE space.
I suppose you want to prove the following statement: For all $\epsilon>0$ there exists $N$ such that for all $n>N$ $$ \|P_nu-u\|_V\le \epsilon \|u\|_K \quad\forall u\in K. $$ Let me argue by contradiction. Suppose the claim does not hold: Then there exists $c>0$, such that for all $n$ there is $u_n\in K$ with $$ \|P_nu_n-u_n\|_V >c \|u_n\|_K . $$ Wlog assume $\|u_n\|_K=1$. Then the set of $u_n$ is bounded in $K$ and pre-compact in $V$ by the compact embedding. And there exists $u\in V$ such that after extracting a subsequence $u_n\to u$ in $V$. Now consider the difference $$ P_n u_n - u = P_n(u_n-u) + (P_n-Id)u. $$ The term $P_nu-u$ vanishes by assumption. By the principle of uniform boundedness the operators $P_n$ are uniformly bounded (stability), hence $P_n(u_n-u)$ vanishes as well. This contradicts the inequality $\|P_nu_n-u_n\|_V>1$. And the claim is proven.