As David Mitra mentioned in his comment, one such proof can be found in
Morrison T.J. Functional Analysis. An Introduction to Banach Space Theory (Wiley, 2000), p.221.
The proof we give is very elementary and avoids the usual argument seen for this fact, which involves either the Baire Category Theorem or the Hahn-Banach Theorem. It is due to the Chinese mathematician Nam-Kiu Tsing (1984).
Proposition 5.1. No infinite-dimensional normed linear space with a countable Hamel basis can be complete.
Proof. Let $(X,\|\cdot\|)$ be a normed linear space with Hamel basis $(e_n)_n$, and note that without loss of generality we can assume $\|e_n\|=1$.
Let $S_{n-1}$ denote the linear subspace of $X$ spanned by $\{e_1,e_2,\dots,e_{n-1}\}$, and let $r_n \equiv \inf\{ \|x+e_n\|; x\in S_{n-1}\}$ for any $n\ge2$.
Since $\theta\in S_{n-1}$, it follows that $r_n\le \|0+e_n\| =1$ for all $n\ge2$. Now since $S_{n-1}$ is finite-dimensional, it is complete, and hence closed in $X$. Since $e_n\notin S_{n-1}$, we have that $r_n>0$ for all $n\ge2$. Now define the following scalar sequence $(t_n)_n$ by $t_1=1$, $t_2=\frac13$ and for $n\ge2$, $t_{n+1}=\frac13r_nt_n$. Then note that we have
$$0<t_{n+k}\le\left(\frac13\right)^k r_nt_n \le \left(\frac13\right)^{n+k-1}$$
for all $n\ge2$ and $k\in\mathbb N$. Now for each $n\in\mathbb N$, define $u_n=\sum_{i=1}^n t_i e_i$ and note that $(u_n)_n$ is a cauchy sequence in $X$.
But also notice that for any element $u=\sum_{i=1}^{m-1} \alpha_i e_i\in X$, we have
\begin{align}
\| u_{n} - u \|
& = \left\|
\sum_{i = 1}^{m - 1} (t_{i} - \alpha_{i}) e_{i} + t_{m} e_{m} +
\sum_{i = m + 1}^{n} t_{i} e_{i}
\right\| \\
& \ge \left\|
\sum_{i = 1}^{m - 1} (t_{i} - \alpha_{i}) e_{i} + t_{m} e_{m}
\right\| -
\left\| \sum_{i = m + 1}^{n} t_{i} e_{i} \right\| \\
& \ge t_{m}
\left\|
\sum_{i = 1}^{m - 1} \frac{1}{t_{m}} (t_{i} - \alpha_{i}) e_{i} + e_{m}
\right\| -
\sum_{i = m + 1}^{n} t_{i}\\
& \ge t_{m} r_{m} -
\sum_{i = 1}^{n - m} \left( \frac{1}{3} \right)^{i} r_{m} t_{m} \\
& \ge \frac{1}{2} t_{m} r_{m}
\end{align}
for all $n>m$. But this means that $\|u_n-u\|$ does not go to zero, which implies $(u_n)_n$ does not converge. Hence, $X$ is not complete.
Tsing N.K. [1984]. Infinite dimensional Banach spaces must have uncountable basis—an elementary proof. Amer. Math. Monthly, 96 (5), 505-506. JSTOR
The symbol $\theta$ is used to denote the zero vector of the space $X$.
Classically, they can be pretty simple: that is,
We can have a model $M$ of ZFC, with an inner model $N$ of ZF, such that there is a $\mathbb{Z}/2\mathbb{Z}$-vector space $V\in N$ such that $(i)$ $N\models$"$V$ has no basis" and $(ii)$ $M\models$"$V\cong\bigoplus_{\omega}\mathbb{Z}/2\mathbb{Z}$".
Of course, inside $N$ this characterization of $V$ won't be visible.
I almost forgot the classic: $\mathbb{R}$, as a vector space over $\mathbb{Q}$! I'd argue this is "more complicated" than the one above in certain senses, but in others its more natural.
As to why this happens: basically, consider a "sufficiently large" vector space $V$ with lots of automorphisms. Then, starting in a universe $M$ of ZFC which contains $V$, we can build a forcing extension $M[W]$, where $W$ is a "generic copy" of $V$. That is, $W$ is isomorphic to $V$, but all twisted around in a weird way. Now, we can take a symmetric submodel $N$ of $M[W]$ - this is a structure between $M$ and $M[W]$, consisting (very roughly) of those things which can be defined from $W$ via a definition which is invariant under "lots" of automorphisms of $W$ - specifically, invariant under every automorphism fixing some finite set of vectors! But as long as $W$ is sufficiently nontrivial, no basis (or, in fact, infinite linearly independent set) is so fixed.
Of course, I've swept a lot under the rug - what's a forcing extension? what exactly is $M[G]$? and why does it satisfy ZF? - but this is a rough intuitive outline.
Actually, in a precise sense, this is the wrong answer: I've just argued that it's consistent with ZF that some vector spaces not have bases. But, in fact, Blass showed that "every vector space has a basis" is equivalent to the axiom of choice! See http://www.math.lsa.umich.edu/~ablass/bases-AC.pdf, which is self-contained. Blass' construction actually proves that "every vector space has a basis" implies the axiom of multiple choice - that from any family of nonempty sets, we may find a corresponding family of finite subsets (so, not quite a choice function); over ZF this is equivalent to AC (this uses the axiom of foundation, though).
Blass argues roughly as follows. Start with a family $X_i$ of nonempty sets; wlog, disjoint. Now look at the field $k(X)$ of rational functions over a field $k$ in the variables from $\bigcup X_i$; there is a particular subfield $K$ of $k(X)$ which Blass defines, and views $k(X)$ as a vector space over $K$. Blass then shows that a basis for $k(X)$ over $K$ yields a multiple choice function for the family $\{X_i\}$.
So now the question, "How can some vector spaces fail to have bases?" is reduced (really ahistorically) to, "How can choice fail?" And for that, we use forcing and symmetric submodels (or HOD-models, which turn out to be equivalent but look very different at first) as above.
Best Answer
Careful: with the axiom of choice we get
$$\dim_{\mathbb{Q}}(\mathbb{R}) = 2^{\aleph_0} = \mathfrak{c}$$
(the cardinality of the continuum). To assert that $2^{\aleph_0} = \aleph_1$ requires the continuum hypothesis.
Without the axiom of choice, if we define the dimension as the size of a basis, then yes, without a basis this particular notion of dimension is undefined. So we might try to use others.
For example, we can talk about the rank, which is the minimal size of a set of generators of $\mathbb{R}$ as a $\mathbb{Q}$-vector space. Since $\mathbb{R}$ is uncountable no countable subset generates. And on the other hand $\mathbb{R}$ itself clearly generates. So assuming the continuum hypothesis the rank is $2^{\aleph_0} = \mathfrak{c}$ again. I don't know if it's consistent with ZF that the rank is somewhere strictly between $\aleph_0$ and $2^{\aleph_0}$.
There is also a somewhat different notion of rank given by the maximal size of a linearly independent subset. We can explicitly exhibit a linearly independent subset of $\mathbb{R}$ of cardinality $2^{\aleph_0}$ in ZF (see this MO answer). So it's provable in ZF that the rank in this sense is $2^{\aleph_0}$.
(In the absence of choice there's a subtle distinction to be made between "minimal size of a set of generators" vs. "size of a minimal set of generators" since without Zorn's lemma a minimal set of generators need not exist, and similarly for "maximal size of a linearly independent subset" vs. "size of a maximal linearly independent subset." Either a minimal set of generators or a maximal linearly independent subset must be a basis. I suppose in the absence of choice cardinals don't need to be totally ordered either though. So either way the rank might end up undefined in general. Welp.)