for each $x,y,z \in X$ we have
\[
\inf_x f_1(x,y) + \inf_z f_2(y,z) \le f_1(x,y) + f_2(y,z)
\]
so the left hand side is a lower bound for $\{f(x,y,z)\mid x,z \in X\}$. As the infimum is the greatest lower bound, we obtain
\[
\inf_x f_1(x,y) + \inf_z f_2(y,z) \le \inf_{x,z} \bigl(f_1(x,y) + f_2(y,z) \bigr).
\]
On the other hand, we have for each $x,y,z \in X$
\[
\inf_{x,z} \bigl(f_1(x,y) + f_2(y,z) \bigr) \le f_1(x,y) + f_2(y,z)
\]
As this holds (fixing $z$ for the moment) for each $x$ we get
\[
\inf_{x,z} \bigl(f_1(x,y) + f_2(y,z) \bigr) \le\inf_x\bigl( f_1(x,y) + f_2(y,z) \bigr) = \inf_x f_1(x,y) + f_2(y,z)
\]
This holds for each $z$ and therefore
\[
\inf_{x,z} \bigl(f_1(x,y) + f_2(y,z) \bigr) \le\inf_z\bigl( \inf_x f_1(x,y) + f_2(y,z) \bigr) = \inf_x f_1(x,y) + \inf_z f_2(y,z).
\]
So the inequality in question holds always, we don't need coninuity or compactness.
Regarding your second question:
Since $X^2$ is compact, $f_1$ is uniformly continuous. Now let $\epsilon > 0$. By uniform continuity there is $\delta > 0$ s. t. $|f_1(x',y') - f_1(x,y)| < \epsilon$ for $d(x',x), d(y',y) < \delta$. Now let $y,y' \in X$ with $d(y',y) < \delta$. Since $X$ is compact, infima of continuous functions are attained and therefore there are $x, x' \in X$ with $g(y) = f_1(x,y)$ and $g(y') = f_1(x',y')$. Assume wlog $g(y) < g(y')$, we have
\begin{align*}
g(y') &= f_1(x',y')\\\
&= \inf_a f_1(a,y')\\\
&\le f_1(x,y')\\\
&\le f_1(x,y) + \epsilon \quad\text{as $d(y,y') < \delta$}\\\
&= g(y) + \epsilon\\\
\iff g(y') - g(y) &\le \epsilon
\end{align*}
So $g$ is (uniformly) continuous.