It is not only that there is, but in fact a vector space is infinitely dimension if and only if it has a proper subspace which is isomorphic to it.
One direction is easy, if the dimension is finite then this is impossible, since the isomorphism is injective and the sum of ranks of the kernel and image implies that the image of the isomorphism has to be everything.
If the space is infinitely dimensional, let $B=\{v_i\mid i\in I\}$ be a Hamel basis of the space, this is an infinite set therefore there is a bijection between $B$ and $B\setminus\{v\}$ for some $v\in B$. This bijection extends unique to an isomorphism (note that the extension is unique, linear, and injective). Clearly the image of such isomorphism is a proper subspace, since $v$ is not in the image.
(Note the axiom of choice was used extensively here for choosing a basis and for the bijection with a proper subset; without the axiom of choice it is consistent that there are very strange vector spaces which are not finitely generated but do not have this property.)