An equivalent formulation of your question is to prove that
$f:X^\prime \to \mathbb{R}$ be written as the supremum of affine functions, i.e.
$f(x^\prime)=\sup_{i \in \mathbb{N}} \langle x^\prime,x_i\rangle+c_i$.
This is equivalent to weak star lower semicontinuity (cf Ambrosio, Fusco, Pallara Functions of Bounded Variation and free discontinuity problems). Perhaps you have some representation that implies this? You don't need the full strength of weakly star lower semicontinuity then (though they are equivalent), since if $x_n^\prime \to x^\prime$ weak star, then we have
\begin{align}
\liminf_n f(x_n^\prime) &\geq \liminf_n \langle x_n^\prime,x_i\rangle+c_i \newline
&=\langle x^\prime,x_i\rangle+c_i
\end{align}
and taking the supremum on the right hand side we obtain sequential lower semicontinuity with respect to weak star convergence, which along with your subsequence implies existence of a minimizer. What does your functional look like?
If you have a bound on the uniform norm, say $\|f\|_{\infty, C}\le M$, the sequence has a uniform Lipschitz estimate $ 2M/r$ on the set $C_r \subset C$ of all points with distance at least $r$ from $\partial C$, so by Ascoli-Arzelà a subsequence does converge uniformly on compact sets of the open sets $C$.
$$*$$
In general, a uniformly bounded sequence $f_n$ of continuous convex functions on a compact convex set $D\subset\mathbb{R}^2$, converging point-wise to a continuous function, need not converge uniformly on $D$. Consider, on the closed unit disk $D$
$$f_n(x,y):=2n^2\Big(x+\frac{y}{n}-1\Big)_+$$
The convex function $f_n$ vanishes on the whole $D$ but a small circular segment $S_n:=\{f_n>0\}$, cut off by the straight line $x+\frac{y}{n}=1$. Note that $\cap_n S_n=\emptyset$. So for any $u\in D$, we have $f_n(u)=0$ eventually. But this convergence to zero is not uniform, because e.g. on the medium point $(x_n,y_n)$ of the arc that bounds $S_n$,
$$x_n:=\frac{n}{\sqrt{n^2+1}},\quad y_n:=\frac{1}{\sqrt{n^2+1}}$$
we have
$$f_n(x_n,y_n):=2n^2\bigg(\sqrt{1+\frac{1}{n^2}}-1\bigg)=1+o(1)\, .$$
Best Answer
Let me resolve a special case of the question, by leveraging the examples mentioned above. Suppose $(K,\rho)$ is a compact metric space and $X$ is the set of Borel probability measures on $K$ endowed with its weak* topology. I claim that $X$ has the convex function property if and only if $K$ is finite.
Theorem 10.2 in "Convex Analysis" by Rockafellar implies that any convex function defined on a finite-dimensional simplex is upper semicontinuous. This gives one direction.
Conversely, suppose $K$ is infinite. Letting $k_\infty$ be an accumulation point of $K$ (which exists because $K$ is an infinite compact metrizable space), define the affine continuous function $\varphi:X\to\mathbb R^2$ given by $\varphi(x):=\int_K \left(\rho(k,k_\infty), \rho(k,k_\infty)^2\right)\text{ d}x(k)$. Then, the convex function $f:X\to\mathbb R$ which takes $f(\delta_{k_\infty}):=0$ and $f(x):=\tfrac{\varphi_1(x)^2}{\varphi_2(x)}$ for every $x\neq\delta_{k_\infty}$ should work. A failure of continuity is witnessed along sequence $(\delta_{k_n})_{n=1}^\infty$ where, $\{{k_n}\}_{n=1}^\infty\subseteq K\setminus\{k_\infty\}$ is a sequence converging to $k_\infty$.
A natural conjecture is now that a general $X$ will have the convex function property if and only if $X$ has finitely many extreme points. Is this true?