Relationship between equicontinuity and total boundedness A subspace of a complete metric space is compact iff it is closed and totally bounded.
Take a compact space $X$, then its space of continuous functions $C(X)$ is complete.
Subspaces of $C(X)$ are compact iff they are closed, bounded and equicontinuous.
For such subspaces, the above two theorems show that closed, bounded and equicontinuous is equivalent to closed and totally bounded.


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*Can we "cancel" closedness in the above equivalence so we have bounded and equicontinuous iff totally bounded?

*Also, we know that totally bounded implies bounded. Does that mean that totally bounded implies equicontinuous in the case of such spaces?
 A: Yes, you can "cancel" closedness. In fact that's often exactly how the Arzela-Ascoli theorem is stated. The statement of Arzela-Ascoli that is taught at my course is as follows:
Let $K$ be a compact space and $C(K)$ be the space of continuous real valued functions on $K$. Then $S \subset C(K)$ is totally bounded if and only if it is bounded and equicontinuous.
In a similar vein rather than the result in your first sentence a slightly different result that talks about non-closed $S$ is: a subset $S$ of a complete metric space is pre-compact (meaning $\bar{S}$ is compact) if and only if $S$ is totally bounded. It's equivalent to the result you've stated since $S$ is totally bounded if and only if $\bar{S}$ is totally bounded.
Finally your second bullet point is correct, a totally bounded subset of $C(K)$ is equicontinuous since it's an if and only if statement.
Proof of Arzela-Ascoli:
Firstly let $S$ be totally bounded. Then certainly $S$ is bounded. To show equicontinuity, fix a point $x \in K$ and $\epsilon > 0$. Then there exists an $\epsilon$-net $\{f_1, \ldots, f_n\} \subset S$. Then since each $f_i$ is continuous there exist an open neighbourhood of $x$, $U_i$, such that for all $y \in U_i$ we have $\lvert f_i(y) - f_i(x) \rvert < \epsilon$.
Now take $U = \cap_{i=1}^n U_i$ which is still an open nbhd of $x$. Then for any $y \in U$ and $f \in S$, there exists $f_i$ such that $\lVert f - f_i \rVert < \epsilon$. Then
$$ \lvert f(y) - f(x) \rvert \leq \lvert f(y) - f_i(y) \rvert + \lvert f_i(y) - f_i(x) \rvert + \lvert f_i(x) - f(x) \rvert < 3 \epsilon $$
So $S$ is equicontinuous.
Conversely let $S$ be bounded and equicontinous. Then by equicontinuity for all $x \in K$ ther exists an open nbhd $U_x$ of $x$ such that for all $y \in U_X$, $f \in S$ we have $\lvert f(x) - f(y) \rvert < \epsilon$. Then $K = \cup_{x \in K} U_x$ so by compactness $K = \cup_{i = 1}^n U_{x_i}$ for some $x_1, \ldots, x_n$. Now consider the map
$$ \varphi: C(K) \to \mathbb R^n, \quad \varphi(f) = (f(x_1), \ldots, f(x_n))$$
It is easy to check that $\varphi$ preserves boundedness of $S$, thus $\varphi(S)$ is a bounded subset of $\mathbb R^k$. Then by Heine-Borel $\overline{\varphi(S)}$ is a compact, thus $\overline{\varphi(S)}$ and hence $\varphi(S)$ is totally bounded in $\mathbb R^k$. Now fix $\epsilon > 0$ and pick an $\epsilon$-net $\{\varphi(f_1), \ldots, \varphi(f_n)\}$ of $\varphi(S)$ in $\mathbb R^k$, we show $\{f_1, \ldots, f_m\}$ is an $3\epsilon$-net of $S$ in $C(K)$.
Fix any $y \in K$ and $f \in C(K)$. Then there exists $f_j$ such that $\lvert f(x_i) - f_j(x_i) \rvert < \epsilon$ for all $i = 1, \ldots, n$. Then $K = \cup_{i=1}^n U_{x_i}$ thus $y \in U(x_i)$ for some $i$. Then
$$ \lvert f(y) - f_j(y) \rvert \leq \lvert f(y) - f(x_i) \rvert + \lvert f(x_i) - f_j(x_i) \rvert + \lvert f_j(x_i) - f(y) \rvert < 3\epsilon$$
Thus $\lVert f - f_j \rVert < 3\epsilon$ so as required $\{f_1, \ldots, f_m\}$ is a $3\epsilon$-net of $S$. So $S$ is totally bounded.
