= Solution
For nonzero finite-dimensional $U$ and $V$, \b[the intended assertion is true]. Write $a=\dim U$, $b=\dim V$, and replace $W$ by $\operatorname{im}\phi$, of dimension $r$. Injectivity on the stated slices implies $\phi(u\otimes v)\ne0$ whenever $u,v\ne0$. Consequently the <bilinear map> defines a continuous map of <Complex projective spaces>
$$
F:\mathbb P(U)\times\mathbb P(V)\longrightarrow\mathbb P(\operatorname{im}\phi),\qquad ([u],[v])\longmapsto[\phi(u\otimes v)].
$$
The rank $r$ is at least one, since one such nonzero tensor has nonzero image.
Let $L_U,L_V,L_W$ be the respective <tautological bundles>. Fiberwise, $\phi$ identifies $L_U\otimes L_V$ with $F^*L_W$. For the positive hyperplane classes $x=c_1(L_U^*)$, $y=c_1(L_V^*)$ and $z=c_1(L_W^*)$ (zero when the projective space is a point), the <first Chern class of a tensor product of complex line bundles> gives
$$
F^*z=x+y.
$$
The <Künneth theorem>, together with part (a), identifies the product <cohomology ring> with
$$
H^*(\mathbb P(U)\times\mathbb P(V);\mathbb Z)=\mathbb Z[x,y]/(x^a,y^b).
$$
There are no Tor terms because the factor groups are free. In particular, its monomials $x^iy^j$ with $0\leq i<a$, $0\leq j<b$ form an integral additive basis. In top degree,
$$
(x+y)^{a+b-2}=\binom{a+b-2}{a-1}x^{a-1}y^{b-1}\ne0.
$$
If $r\leq a+b-2$, the target relation $z^r=0$ would imply $(x+y)^r=0$ and hence contradict this nonzero top power. Thus
$$
\boxed{\dim_{\mathbb C}\operatorname{im}\phi\geq\dim_{\mathbb C}U+\dim_{\mathbb C}V-1.}
$$
When $a=b=1$, the already-established inequality $r\geq1$ gives the same conclusion. The <complex bilinear dimension bound> is sharp: multiplication of complex polynomials of degrees less than $a$ and $b$ has target dimension $a+b-1$, is injective in either nonzero fixed factor, and its image spans every monomial in that target.
\b[Literal zero-space qualification.] The printed assertion does not explicitly exclude zero vector spaces. If $U=0$, $V=\mathbb C^2$ and $W=0$, its slice-injectivity hypothesis is vacuous, while the claimed inequality would be $0\geq1$. Thus, with zero spaces permitted, this is a counterexample to the assertion exactly as written; the proof above supplies the usual nonzero finite-dimensional interpretation.
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