= Solution
The rank assumption says that $\Gamma'=\Gamma\cap V'$ is a full lattice in the real vector space underlying $V'$. Thus $X'=V'/\Gamma'$ is compact, and the inclusion $V'\hookrightarrow V$ descends to an injective holomorphic homomorphism $X'\hookrightarrow X$. It is therefore a <complex subtorus>.
Conversely, let $j:Y\hookrightarrow X$ be a subtorus. Its differential at the identity identifies the universal cover of $Y$ with a complex subspace $V'\subseteq V$. Lifting $j$ to universal covers shows that the period lattice of $Y$ is precisely $V'\cap\Gamma$. Compactness of $Y$ makes this a full lattice of rank $2\dim_{\mathbb C}V'$, so every subtorus has the stated form.
Now let $H$ be a polarisation on $X$. Its restriction to $V'$ is still positive definite, and its imaginary part remains integral on $\Gamma'$, so it polarises $X'$. Define the Hermitian orthogonal complement
$$
V''=(V')^{\perp_H}.
$$
Because $V'$ is spanned over $\mathbb R$ by lattice vectors and $E=\operatorname{Im}H$ is integral on $\Gamma$, the real equations $E(v,\gamma')=0$ for $\gamma'\in\Gamma'$ make $V''$ rational with respect to $\Gamma$. Hence $\Gamma''=V''\cap\Gamma$ is a full lattice in $V''$, and $X''=V''/\Gamma''$ is a subtorus. Since $V=V'\oplus V''$, the lattice $\Gamma'+\Gamma''$ has finite index in $\Gamma$. Consequently the addition map
$$
X'\times X''\longrightarrow X
$$
is an <isogeny of complex tori>: it is surjective and has finite kernel. Therefore $X=X'+X''$ and $X'\cap X''$ is finite.
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