Consider the case where $S$ has nonempty boundary in $T$ that intersects $S$, and let $H=S$. Then trivially $\partial_S H = \partial_S S = \emptyset$, but $\partial_T S\cap S \neq\emptyset$.
For example, take $S=[0,1]$ in $T=\mathbb{R}$ and $H=[0,1]$. Then $\partial_T H = \partial_T([0,1]) = \{0,1\}$.
However, $\partial_S H = \emptyset$, because no $S$-neighborhood of $H$ intersects the complement of $H$ in $S$.
If you want an example with $H\neq S$, then just take $H=(0,1]$ and $S$, $T$ as before. Then $\partial_T(H)\cap S = \{0,1\}$, but $\partial_S(H) = \{0\}$.
On the other hand, it is always true that $\partial_S H\subseteq \partial_T H\cap S$:
Let $x\in \partial_S H$, and let $U$ be an open set of $T$ that contains $x$. Then $U\cap S$ is open in $S$ and contains $x$, hence
$$\varnothing \neq H\cap (U\cap S) = (H\cap S)\cap U = H\cap U.$$
So $H\cap U\neq\varnothing$. And
$$\varnothing \neq (S-H)\cap (U\cap S) \subseteq (T-H)\cap U,$$
hence $U\cap (T-H)\neq \varnothing$. Hence $x\in \partial_T H$, as claimed.
You should be really careful with the complement: note that you have two different notions of "$H^c$" at play in the situation above: there's the complement of $H$ in $S$, and the complement of $H$ in $T$. It's probably better to use $S-H$ and $T-H$ in your arguments, to avoid possible confusion (though, of course, $S-H\subseteq T-H$).
Looks fine, for the most part, though I have no idea what you could mean by $U\cap A$ if $U$ is an open cover of $A$. I suspect that you instead mean that $C$ is a compact subset of $X$ containing an open neighborhood $U$ of $x.$ Also, closed subsets of Hausdorff spaces need not be compact (consider $\Bbb R$ as a subset of itself, for example), though compact subsets of Hausdorff spaces will be closed.
Alternately, you can use your previous result to proceed even more directly in showing that $C\cap A$ is compact. Since $A$ is closed, then $C\cap (X\setminus A)$ is open in $C,$ so $C\cap A=C\setminus \bigl(C\cap(X\setminus A)\bigr)$ is closed in $C,$ so is compact by your previous result since $C$ is compact.
Best Answer
Suppose that $H$ is a closed subspace of $X$, and $F$ is closed in the subspace $H$. By definition of the relative topology there is a closed set $C$ in $X$ such that $F=C\cap H$. But then $F$ is the intersection of two closed subsets of $X$, so $F$ is closed in $X$.
If you don’t already know this characterization of closed sets in the relative topology, it’s worth proving as a separate
Of course for this you do need to look at complements, but it’s very easy. $Y\setminus H$ is open in $Y$, so there is an open $U\subseteq X$ such that $Y\setminus H=U\cap Y$. Let $F=X\setminus U$; then $F$ is closed in $X$, and $F\cap Y=(X\setminus U)\cap Y=Y\setminus U=H$.