$\def\conv{\mathop{\mathrm{conv}}}$Recall that $\conv X$ is the set of all convex combinations of points from $X$. In a convex combination 
$$
  f=\sum_{i=1}^k\alpha_ix_i, \qquad x_i\in X, \quad \alpha_i>0,
$$
for a point $f\in F$, all the points $x_i$ should lie in $F$ (and hence in $X\cap F$); indeed, if, say, $x_k\notin F$, then $f$ is a relatice interior point of the segment between $x_k$ and
$$
  \frac1{1-\alpha_k}\sum_{i=1}^{k-1}\alpha_ix_i
$$
which is impossible.
Thus $f\in\conv(X\cap F)$.