I make some additional assumptions and simplify notations, nothing of which should cause confusion. Suppose for simplicity that data is generated according to $Y = \beta X + \epsilon$, where all variables are $\mathbb R-$valued and $\epsilon$ has zero mean and variance $\sigma^2$. Assume $X$ has the necessary moments. We have $n$ independent copies of the pair $(Y, X)$; $x = [x_1, \dots, x_n]'$ and $y = [y_1,\dots, y_n]'$.
The OLS estimator of $\beta$ is
\begin{align}
\hat{\beta} &=y'x / x'x \\
&= (x\beta + e)'x/x'x \\
& = \beta + \frac{e'x}{x'x} \\
&= \beta + \frac{\frac{1}{n}\sum_i \epsilon_i x_i}{\frac{1}{n}\sum_i x_i^2}
\end{align}
where $e = [\epsilon_1, \dots, \epsilon_n]'$. The assertion that this approaches $\beta + {\rm Cov}(X, \epsilon) / {\rm Var}(X)$ as $n\to\infty$ is usually in the sense of convergence in probability. According to standard definitions, an estimator is consistent if it converges in probability to the true parameter value, i.e. in this case if $\hat{\beta} \to \beta$ in probability. Here, we had an extra term, in general non-zero, on the right hand side so the estimator is inconsistent.
On the other hand, we say that $\hat{\beta}$ is unbiased if $\mathbb E \hat{\beta} = \beta$. This statement has nothing to do with convergence. All the same, the expectation of the right hand side is again $\beta$ + some term which is not zero in general. Thus, $\hat{\beta}$ is also biased. This answers the first question.
Regarding the second question, notice that bias is usually regarded as a property of estimators, and $\beta$ is unknown so it's not an estimator. Therefore, it does not make sense to speak of the bias of $\beta$. If we are liberal in the usage of bias and let it apply to anything, we see that $\mathbb E \beta = \beta$ so it would be "unbiased".
Without further assumptions on the dependence between $X$ and $\epsilon$ there really isn't any way to tell what the answer to three is, I believe. I should say I did not have time to go through the calculations so I may be wrong.
In light of the discussion under the other answers: If one declares a new definition of consistency in terms of the variance approaching zero, I don't see how inconsistency under the assumption ${\rm Cov}(X, \epsilon) \neq 0$ can be either confirmed or disproved without more assumptions. It's also important to note that consistency in the standard definition says nothing about decreasing variance or decreasing bias. These are separate concepts and should not be mixed up.
You are right that the conditional variance is not generally the same as the unconditional one. By the variance decomposition lemma, which says that, for r.v.s $X$ and $Y$
$$
Var(X)=E[Var(X|Y)]+Var[E(X|Y)]
$$
Translated to our problem,
$$
Var(\widehat{\beta})=E[Var(\widehat{\beta}|X)]+Var[E(\widehat{\beta}|X)]
$$
Now, using that OLS is conditionally unbiased (under suitable assumptions like exogeneity assumed here), we have
$$
E(\widehat{\beta}|X)=\beta
$$
and thus
$$
Var[E(\widehat{\beta}|X)]=0,$$
as $\beta$ is a constant, so that
$$
Var(\widehat{\beta})=E[Var(\widehat{\beta}|X)]
$$
or
$$Var(\widehat{\beta})=E[\sigma^2(X'X)^{-1}]=\sigma^2E[(X'X)^{-1}].$$
Best Answer
I find the most difficult parts of this subject to be in keeping track of what is is being integrated over (the expectation with respect to what?). Many expositions take this lightly, and the end result can be highly confusing. Here is The Elements of Statistical Learning on this point
I think you're seeing a bit of that in the wikipedia exposition.
My Preferred Derivation
Here's a detailed break down. Working out the exposition in this way is what finally cleared up the various concepts at play for me. I hope it helps you too.
The first step is to condition on $x$, and take the expectation with respect to $y$. There are no assumptions for this part, $g$ is a general function
$$\begin{align*} ESE(f; x) &= E_Y \left[ \left( y - g(x) \right)^2 \mid x \right] \\ & = E_Y \left[ \left( y - E[y \mid x] + E[y \mid x] - g(x) \right)^2 \mid x \right] \\ & = E_Y \left[ \left( y - E[y \mid x] \right)^2 \mid x \right] + E_Y \left[ \left(E[y \mid x] - g(x) \right)^2 \mid x \right] \\ & \quad + 2 E_Y \left[ \left( y - E[y \mid x] \right) \left( E[y \mid x] - g(x) \right) \mid x \right] \\ \end{align*}$$
The trick here is introducing the $E[ y \mid x ]$ into the middle of the sum. This expectation is over $y$ (the only thing that makes sense in this context), so $E[ y \mid x ]$ is just a function of $x$, same as $g$.
Now we may observe that in
$$ 2 E_Y \left[ \left( y - E[y \mid x] \right) \left( E[y \mid x] - g(x) \right) \mid x \right] $$
the second factor has no dependence on $y$, and the first factor is zero in expectation, so this term dies, and we are left with
$$\begin{align*} ESE(f; x) = E_Y \left[ \left( y - E[y \mid x] \right)^2 \mid x \right] + E_Y \left[ \left(E[y \mid x] - g(x) \right)^2 \mid x \right] \\ \end{align*}$$
The first term is the irreducible error of the classifier. It's your $\sigma^2$ above.
Its worth noting that only the integrand of the first term in this result has any dependence on $y$, so the $E_Y$ may be dropped from the second term.
To get to the bias-variance breakdown, we introduce an expectation over the sampling distribution of $x$. That is, we consider $g$ itself as a random variable; training a learning algorithm takes us from a data set $D$ to a function $g$. Thus, it makes sense to take expectations over this sampling distribution with $g$ as a random variate. I'll write $g(x, D)$ to emphasize this dependence of $g$ on the training dataset $D$. With this notation, the second term in our final equation above becomes
$$ \left[ \left(E[y \mid x] - g(x,D) \right)^2 \mid x \right] $$
Taking the expectation of everything over $D$, and writing $Eg(x)$ for the expectation of $g(x, D)$ with respect to this sampling distribution, we can break this down further
$$\begin{align*} & E_{D} \left[ \left(E[y \mid x] - g(x,D) \right)^2 \mid x \right] \\ & = E_{D} \left[ \left(E[y \mid x] - Eg(x) + Eg(x) - g(x,D) \right)^2 \mid x \right] \\ & = E_{D} \left[ \left(E[y \mid x] - Eg(x) \right)^2 \mid x \right] + E_{D} \left[ \left(Eg(x) - g(x, D) \right)^2 \mid x \right] \\ & \quad + 2 E_{D} \left[ \left(E[y \mid x] - Eg(x) \right) \left( Eg(x) - g(x,D) \right) \mid x \right] \\ \end{align*} $$
The same trick applies. In this breakdown the first factor has no dependence on $D$, and the second is zero in expectation, so the cross term dies. Therefore
$$\begin{align*} E_{D} \left[ \left(E[y \mid x] - g(x,D) \right)^2 \mid x \right] = E_{D} \left[ \left(E[y \mid x] - Eg(x) \right)^2 \mid x \right] + E_{D} \left[ \left(Ef(x) - g(x, D) \right)^2 \mid x \right] \end{align*}$$
The first term here is the bias, and the second is the variance.
Of course, I kept the conditioning on $x$ the entire time, so what we have really arrived at is the pointwise decomposition. To get the usual full decomposition of the total error, all that is needed is to integrate out $x$ as well.
How This Relates to the Wikipedia Breakdown
Wikipedia writes $f(x)$ for my $E[ y \mid x ]$. I think it makes the exposition easier to follow to make this explicit. Wikipedia writes
$$ y = f(x) + \epsilon $$
with the assumption that $E(\epsilon) = 0$. What is really meant is $E(\epsilon \mid x) = 0$. This is implicit in my exposition, I simply use $E[y \mid x]$ throughout.
Note that no independence assumptions are necessary for the bias variance decomposition. What wikipedia says
what they really mean is that $\hat{f}$ can be considered a constant when conditioning on $x$, so it can be taken out front of the expectation.