Let $B_i$ be Hilbert spaces with the continuous inclusions
$B_1 \rightarrow B_2 \rightarrow B_3.$
further assume that $B_1$ is compactly imbedded in $B_2$. Then I need to show that given any $\epsilon > 0$ there exists $C_{\epsilon} >0$ such that for all $u \in B_1$
$||u||_{B_2} \leq \epsilon||u||_{B_1}+C_{\epsilon}||u||_{B_3}.$
Rewritten more clearly on the request of @am_11235... . I thank them for their assistance in helping improve the answer to this question. The original answer was incorrect.
This question is actually exercise 6.12 of Brezis' Functional Analysis. It is attributed to J.L. Lions.
We work by contradiction. Suppose for an $\epsilon>0$, we have that no $C_\epsilon$ works. Then, we have a sequence $v_m \in B_1$ such that $$\|v_m\|_{B_2} >\epsilon \|v_m\|_{B_1} + m\|v_m\|_{B_3}$$ (contradict the statement for all natural number values of $C_\epsilon$). Observe that $v_m \neq 0$ for all $m$ (as an element of $B_2$), because such an inequality cannot hold for the zero vector.
Let $u_m = \frac{v_m}{\|v_m\|_2}$. If we divide the above inequality by $\|v_m\|_2$, we get $$ 1 >\epsilon\frac{\|v_m\|_{1}}{\|v_m\|_{2}} + m\frac{\|v_m\|_{B_3}}{\|v_m\|_{B_2}} $$ Which becomes $$ 1 >\epsilon \|u_m\|_{B_1} +m \|u_m\|_{B_3} $$ with the knowledge that $||u_m||_{B_2} = 1$ for all $m$.
(The error in the previous edit was the normalization I used here. It led to some typos down the line).
Observe that $\|u_m\|_{B_1} < \frac 1{\epsilon}$ for all $m$. Therefore, $\{u_m\}$ is a bounded sequence in $B_1$.
As the inclusion map $B_1 \to B_2$ is compact, there exists a convergent subsequence in $B_2$, given by $u_{m_k}$. Say that $u_{m_k} \to u$ in $B_2$.
As the inclusion map from $B_2$ to $B_3$ is continuous, one sees that $u_{m_k} \to u$ in $B_3$ as well.
However, let's go back to the inequality with $u_{m_k}$ used instead of $u_m$ : $$ 1 >\epsilon \|u_{m_k}\|_{B_1} +m_k\|u_{m_k}\|_{B_3} $$
Now, if $u \neq 0$, then $\|u\| \neq 0$. In particular, we can find $K$ larger enough such that $\|u_{m_k}\| > \frac{\|u\|}{2}$ for all $k>K$. But then, $m_k\|u_{m_k}\|_{B_3} > m_k \frac{\|u\|}{2} \to \infty$ as $k \to \infty$. This contradicts the inequality.
It follows that $u = 0$.
However, we also had $u_{m_k} \to u$ in $B_2$. This is a problem, since $\|u_{m_k} - 0\| = \|u_{m_k}\| = 1$ for all $k$, so there is no way that $u_{m_k} \to 0$ in $B_2$.
Finally, the contradiction is complete, and the sequence of $u_k$ cannot exist. Thus, there is a $C_{\epsilon}>0$ such that the proposed inequality holds for all $u$.