What does "admissible" mean for you? Does it imply smoothness (stabilisers are open)? If not then I think the statement might be false (choose some hopelessly discontinuous injection from $\mathbf{Q}_p$ into $\mathbf{C}$ and then consider the induced "natural" representation of $GL(2,\mathbf{Q}_p)$ on $\mathbf{C}^2$; that's definitely irreducible, and the space of vectors fixed by some compact open will definitely be finite-dimensional).

Assuming you want to assume smoothness, you're also still a bit out, because there are infinitely many finite-dimensional smooth irreducible admissible representations of $GL(2,\mathbf{Q}_p)$---but they are all 1-dimensional. The proof is: if $V$ is such a thing, then choose a basis for $V$ and for each basis vector choose a compact open subgroup stabilising it. The intersection of these guys is still compact and open, and fixes everything. So the kernel of the representation contains a compact open subgroup. But this is a bit worrying because the kernel is normal. Now use the fact that the normal subgroup generated by matrices $(1,e;0,1)$ and $(1,0;e,1)$ for $e$ small is still the whole of $SL(2,\mathbf{Q}_p)$ to deduce that $SL(2,\mathbf{Q}_p)$ is in the kernel, and now the action has to factor through the abelianisation of $GL(2,\mathbf{Q}_p)$. But now Schur's Lemma, which works for smooth irreducible representations, says $V$ is 1-dimensional. It can be any of the uncountably many characters of $\mathbf{Q}_p^\times$ though.

What does "admissible" mean for you? Does it imply smoothness (stabilisers are open)? If not then I think the statement might be false (choose some hopelessly discontinuous injection from $\mathbf{Q}_p$ into $\mathbf{C}$ and then consider the induced "natural" representation of $GL(2,\mathbf{Q}_p)$ on $\mathbf{C}^2$; that's definitely irreducible, and the space of vectors fixed by some compact open will definitely be finite-dimensional).

But assuming smoothness too, it is true that any finite-dimensional smooth irreducible representation of $GL(2,\mathbf{Q}_p)$ is 1-dimensional. The proof is: if $V$ is such a thing, then choose a basis for $V$ and for each basis vector choose a compact open subgroup stabilising it. The intersection of these guys is still compact and open, and fixes everything. So the kernel of the representation contains a compact open subgroup. But this is a bit worrying because the kernel is normal. Now use the fact that the normal subgroup generated by matrices $(1,e;0,1)$ and $(1,0;e,1)$ for $e$ small is still the whole of $SL(2,\mathbf{Q}_p)$ to deduce that $SL(2,\mathbf{Q}_p)$ is in the kernel, and now the action has to factor through $GL(2,\mathbf{Q}_p)/SL(2,\mathbf{Q}_p)=\mathbf{Q}_p^\times$. But now Schur's Lemma, which works for smooth irreducible representations, says $V$ is 1-dimensional.