Given the axiom of choice, every vector space has a basis (though it will be a very unnatural basis), and you are correct that infinite-dimensional vector spaces are exactly those where the basis is infinite.
But this kind of basis (often called a Hamel basis) is rather useless and impossible to visualize.
So, a more concrete way of thinking about it might be that in an infinite-dimensional vector space, you can exhibit infinitely many vectors $v_1, v_2, v_3, \ldots$ that are all linearly independent; no (finite) linear combination of vectors is zero. Equivalently, $v_n$ is not a linear combination of $v_1, v_2, \ldots, v_{n-1}$ for any $n$.
In particular, this means $\oplus_{i=1}^\infty \mathbb{R}$ (the set of infinite sequences of real numbers where all but finitely many terms are zero) is algebraically a subspace of every infinite-dimensional vector space over $\mathbb{R}$.
How do such vector spaces differ from finite-dimensional vector spaces? Many things break. For example:
Some linear maps do not have any eigenvalues.
Some linear maps are not continuous; you end up having to restrict to Bounded operators which are basically, operators which behave nicely with the norm on your vector space.
The Dual space of the dual space of $V$, $(V^*)^*$, might not equal $V$.
You are wise to be cautious when working over infinite dimensions, since not everything that's true for finite dimensional vector spaces is necessarily true for infinite dimensions. For example, this problem says that the Heine-Borel theorem can fail in infinite dimensions.
That being said, you're being a little too cautious. The slick arguments you were concerned about do work:
Given any $e_k\in A$, $(e_k)_n=1$ when $n=k$ and $0$ otherwise, so it is indeed true that $\|e_k\|=1$. This is sufficient to show $A$ is bounded.
$A$ is trivially closed because each point in $A$ is isolated. Indeed, simply consider the the open sets you define in (d). Put differently, you can say that there can be no sequence $\{a_n\}\subset A$ convergent to a point outside of $A$ since such a convergent sequence would also be Cauchy (we're working in a metric space). Since this sequence $\{a_n\}$ doesn't converge to a particular $e_j$, for any $N$ there is some $n> N$ such that $\|a_n-e_j\|=\sqrt{2}>1$ (since $a_n=e_i$ for some $i\neq j$). This is true for all $j$, so it is not hard to show from here that $\{a_n\}$ is not Cauchy: Let $\epsilon>0$ be smaller than $1$ and let $N$ be any positive integer. Then there is some $n$ with $\|a_n-a_N\|>1$, so it is not true that $n,m\geq N$ implies $\|a_n-a_m\|<\epsilon$.
This also provides a slick proof for (d), since a subset of a metric space is compact if and only if every sequence has a convergent subsequence. In particular, the sequence of the basis vectors $\{e_n\}$ has no convergent subsequences since such a subsequence would be Cauchy, but each point in the sequence is distance $1$ from each other.
That being said, your argument for (d) is mostly fine, but the notation can be improved. When you suppose for contradiction that $A$ is compact, we don't know which $U_i$ we're pulling into our finite subcover--we only know there are finitely many (writing $U_i$ means something! It is the particular open set containing $e_i$!). It would be better to say that our finite subcover has the form $\{U_{i_1},\dotsc, U_{i_m}\}$ for some positive integers $i_1,\dotsc, i_m$. Then take $n>\max(i_1,\dotsc, i_m)$, and note that $e_n$ isn't covered, as you have already done.
Best Answer
A compact set in a metric space must be bounded. Otherwise we can take $\{ x_n \}_{n=1}^\infty$ and a fixed point $x_0$ such that $d(x_n,x_0) \geq n$. This will have no convergent subsequence, which we can prove by showing that it has no Cauchy subsequence.
A compact set in a metric space (also in a Hausdorff space) must be closed. Otherwise we can pick a sequence which converges to a point in the closure which is not in the set. This will have no convergent subsequence (within the set, at least).
A closed and bounded set in a metric space need not be compact. In an infinite dimensional Banach space, closed balls are not compact. For example, in $\ell^p$, $\{ e_1,e_2,\dots,e_n,\dots \}$ is a sequence in the closed unit ball which has no convergent subsequence. In an infinite dimensional Hilbert space we can essentially copy this example by finding an orthonormal sequence. The result holds in any infinite dimensional Banach space, but the details of the construction are not quite so trivial.
A compact subset of an infinite dimensional Banach space can be infinite dimensional, in the sense that it is not contained in any finite dimensional subspace. One way to generate infinite dimensional compact sets is to ensure that any sequence of linearly independent vectors converges to zero. So for example, if $X=\ell^p$, consider the set $\{ 0,e_1,e_2/2,\dots, e_n/n,\dots \}$. This is clearly infinite dimensional, but it is also compact, since any sequence in this set either converges to $0$ or has a constant subsequence.
In general, a subset $A$ of a metric space $X$ is compact if and only if it is complete and totally bounded. A subset of a metric space is totally bounded if for any $\varepsilon > 0$ there exists $N \in \mathbb{N}$ and points $x_1,\dots,x_N$ such that $A \subset \bigcup_{n=1}^N B(x_n,\varepsilon)$. In a Banach space $X$, this is equivalent to saying that it is bounded and, for every $\varepsilon > 0$ there exists $N \in \mathbb{N}$ and a subspace $V$ of $X$ whose dimension is $N$ such that $(\forall x \in A) \, d(x,V) < \varepsilon$.