Let $H,K$ be non-zero Hilbert spaces. Then the algebraic tensor product $H \odot K$ is an inner product space for the unique inner product determined by $$\langle h \otimes k, h' \otimes k'\rangle = \langle h, h'\rangle_H \langle k, k' \rangle_K$$
I am trying to prove that $H \odot K$ is a Hilbert space (i.e. $H \odot K$ is complete for the norm induced by this inner product) implies that $\dim H <\infty$ or $\dim K < \infty$.
Let us prove the contrapositive, i.e. so suppose that $H$ and $K$ are both infinite dimensional. How can I show that there is a Cauchy sequence in $H \odot K$ that does not converge?
It is very easy to exhibit a non-converging Cauchy sequence in $H\odot K$, but actually proving that it does not converge is a bit more difficult. The approach suggested by @Max, although somewhat sophisticated, is perhaps the easiest way to pin down all of the details. Here is a pedestrian way to describe this method.
For each pair of vectors $(x, y)\in H\times K$, consider the bounded linear operator $$ T_{x, y}:H\to K $$ given by $$ T_{x, y}(z) = \langle x,z\rangle y, \quad\forall z\in H. $$ (For simplicity I am assuming that we are working with real vector spaces but this can be fixed in the complex case by seeing $T_{x, y}$ as an operator defined on $\bar H$, namely the Hilbert space made out of $H$ by adopting the new scalar multiplication operation $\lambda \cdot x:= \bar \lambda x$).
It is easy to see that the map $$ (x, y)\in H\times K\mapsto T_{x, y}\in \mathscr {B}(H,K) $$ is bilinear so, by the universal property of (algebraic) tensor products, there exists a unique linear map $$ T :H\odot K\to \mathscr {B}(H,K), $$ such that $T (x\otimes y) = T_{x,y}$. We next claim that $$ \langle T (\xi )x,y\rangle = \langle \xi ,x\otimes y\rangle , \tag {1} $$ for every $\xi $ in $H\odot K$, $x\in H$, and $y\in K$. To prove this write $\xi =\sum_{i=1}^nx_i\otimes y_i$, and notice that $$ \langle T (\xi )x,y\rangle = \sum_{i=1}^n \langle T_{x_i,y_i}(x),y\rangle = $$ $$ = \sum_{i=1}^n \langle \langle x_i,x\rangle y_i,y\rangle = \sum_{i=1}^n \langle x_i,x\rangle \langle y_i,y\rangle = $$ $$ = \sum_{i=1}^n \langle x_i\otimes y_i,x\otimes y\rangle = \langle \xi ,x\otimes y\rangle, $$ proving the claim.
Note that the range of each $T_{x,y}$ is the one-dimensional space spanned by $y$, so $T_{x,y}$ has rank 1. Furthermore, every $\xi $ in $H\odot K$ may be writen as $\xi =\sum_{i=1}^nx_i\otimes y_i$, so $$ T (\xi ) = \sum_{i=1}^n T_{x_i,y_i}, $$ so we see that $T (\xi )$ has rank at most $n$, hence finite, for every $\xi $ in $H\odot K$.
The final contradiction will be achieved by proving that, if $H$ and $K$ are infinite dimensional, and $H\odot K$ is complete, there exists a vector $\xi $ in $H\odot K$ such that $T (\xi )$ has infinite rank.
Let us therefore assume from now on that $H$ and $K$ are both infinite dimensional, so we may find (necessarily infinite) orthonormal bases $\{e_i\}_{i\in I}$, and $\{f_j\}_{j\in J}$, for $H$ and $K$ respectively, and it is well known that $\{e_i\otimes f_j\}_{(i, j)\in I\times J}$ is an orthonormal basis for $H\otimes K$.
Assuming by contradiction that $H\odot K$ is complete, observe that $H\odot K$ coincides with its completion, so $H\odot K = H\otimes K$.
Choose infinite subsets $\{i_n:n\in {\mathbb N}\}\subseteq I$, and $\{j_n:n\in {\mathbb N}\}\subseteq J$, and let $$ \xi = \sum_{n=1}^\infty {1\over n^2} (e_{i_n}\otimes f_{j_n}). $$ This is well defined because the series is clearly absolutely convergent and we are working in a complete space!
Using (1) we have that $$ T (\xi )e_{i_n} = \sum_{j\in J} \langle T (\xi )e_{i_n}, f_j\rangle f_j = \sum_{j\in J} \langle \xi , e_{i_n}\otimes f_j\rangle f_j = \langle \xi , e_{i_n}\otimes f_{j_n}\rangle f_{j_n} = {1\over n^2} f_{j_n}. $$ This shows that ${1\over n^2} f_{j_n}$ lies in the range of $T (\xi )$, and hence that $T (\xi )$ has infinite rank, a contradiction.