First of all, let's see what Mejia's answer says:
Recall that given two topologies $T_1, T_2$ on a set $X$, one has $T_1\subset T_2$ if and only if $T_2$ is finer than $T_1$, if and only if the identity map $id: (X,T_2)\to (X,T_1)$ is continuous, if and only if the identity map $ (X,T_1)\to (X,T_2)$ is closed (i.e. maps closed subsets to closed subsets).
Now, what Mejia explains is that if $T_1\subset T_2$ and $K\subset (X,T_2)$ is compact, then $K\subset (X,T_1)$ is also compact. Of course, one does not need to quote Mejia on this: The observation is a special case of the fact (which you surely learned the first topology course) that the image of a compact under a continuous map is again compact.
Next, in the 2nd paragraph of your post you say:
If a set is compact under the topology of uniform convergence, then it is compact under the topology of pointwise convergence.
Let's make sense of this statement. First of all, you should fix a set $A$ and a metric space $(B,d)$, and consider two topologies on the set $M=Map(A,B)=B^A$ of all (possibly discontinuous) functions $A\to B$. The first topology, $T_1$, is the "topology of pointwise convergence" on $B^A$, which is simply the product topology (see here). Note that this topology is not 1st countable.
The second topology, $T_2$, on $M$ is the uniform topology (also known as the "topology of uniform convergence"), it is induced by the metric on $M$:
$$
d(f,g)= \min \{1, \sup_{a\in A} d(f(a), g(a))\}, f, g\in M.
$$
(Note that the supremum can be infinite, hence, the need to take the $\min$.)
What you correctly observed is that the uniform topology $T_2$ is finer than the product topology $T_1$ on $B^A$. Hence, we do have:
Fact 1. If $K\subset (M, T_2)$ is compact, then $K$ is compact in $(M, T_1)$.
Your next guess:
If a set is sequentially closed under the topology of pointwise convergence, then it is sequentially closed under the topology of uniform convergence.
is correct even without the adjective "sequential": It is simply:
Fact 2. The identity map $(M, T_1)\to (M,T_2)$ is closed (see the 1st paragraph above).
Now, in the last paragraph, you say:
However, sequential closeness is a necessary condition for compactness.
Again, there is no need to say "sequential" here: You are simply stating:
Fact 3. Each compact subset of a Hausdorff topological space is closed. (If you insist on "sequential" statements, it becomes Each sequentially compact subset of a Hausdorff topological space is sequentially closed.)
From this, you somehow conclude that
above two results look "contradictory"
(I assume, the "two results" are Facts 1, 2 and 3 above.)
Now, what is "contradictory" here? Take a compact subset $K\subset (M, T_2)$, then it is compact in $(M,T_1)$ (Fact 1), hence, it is closed in $(M,T_1)$ (Fact 3). But it is also closed in $(M,T_2)$, according to Fact 2 (since $K$ was compact in $(M,T_2)$ to begin with).
There is no contradiction here whatsoever.
Maybe you mean to take a compact $K\subset (M,T_1)$. Of course, then Fact 1 says nothing at all, Fact 2 says that $K$ is closed in $(M,T_2)$ (since $K$ is closed in $(M,T_1)$ due to its assumed compactness). It is also closed in $(M,T_1)$, due to its compactness. So, fine, you discovered that the identity map $(M, T_2)\to (M,T_1)$ does send some closed subsets to closed subsets (specifically, compact subsets); this does not contradict the fact that the identity map $(M, T_2)\to (M,T_1)$ is not closed.
Aside: If $A$ is a topological space, then the subset of continuous functions $C(A,B)$ is closed in $(M, T_2)$ but not in $(M,T_1)$. This is just one of the many manifestations of the fact that the identity map $(M, T_2)\to (M,T_1)$ is not closed. As you know, $C(A,B)$ is not compact in $(M, T_2)$, so, again, there is no contradiction with Fact 1.
All in all, we did not discover any contradictions. Maybe Facts 1, 2, 3 "look contradictory" (when applied to the product and uniform topologies) if you do not look at them closely...
