Properties of the sample mode for Bernoulli data Suppose we have a sample $X_1,...,X_n \sim \text{IID Bern}(p)$ of Bernoulli values with probability parameter $p \neq 0.5$.  Denoting the sample proportion $\hat{p}_n$ we define the sample mode as:
$$\widehat{\text{mode}}(\mathbf{x}_n) = \begin{cases}
0 & & \text{if } \hat{p}_n < \tfrac{1}{2}, \\[6pt]
\tfrac{1}{2} & & \text{if } \hat{p}_n = \tfrac{1}{2}, \\[6pt]
1 & & \text{if } \hat{p}_n > \tfrac{1}{2} \\[6pt]
\end{cases}$$
It's easy to see (and show) that this is a consistent estimator of the true mode of the distribution, but I'm unsure on how to 1) prove it is biased, and 2) derive an expression for the MSE.

Fix the sample size $n$ and let $\hat p_n$ be any function of the data whatsoever.
The data are modeled as $n$ independent Bernoulli$(p)$ variables $X_1,\ldots, X_n.$
Suppose $\hat p_n$ gives an unbiased estimator of the mode--let's call it $\tau.$ To state what that means, notice that $\tau$ partitions all possible values of the variables into three (disjoint) sets:
$$\begin{aligned}
\mathcal{E}_0 = \{(x_1,x_2,\ldots,x_n)\mid \hat p_n(x_1,\ldots,x_n) \lt \frac{1}{2}\} &; \quad \tau = 0\\
\mathcal{E}_{1/2} = \{(x_1,x_2,\ldots,x_n)\mid \hat p_n(x_1,\ldots,x_n) = \frac{1}{2}\} &; \quad\tau=\frac{1}{2} \\
\mathcal{E}_1 = \{(x_1,x_2,\ldots,x_n)\mid \hat p_n(x_1,\ldots,x_n) \gt \frac{1}{2}\}; &\quad\tau = 1.\end{aligned}$$
The expectation is the sum of the values of $\tau$ times their chances:
$$\begin{aligned}
Q(p) = E[\tau(X_1,\ldots, X_n);p] &= 0\times \Pr(\mathcal{E}_0; p) + \frac{1}{2} \times \Pr(\mathcal{E}_{1/2}; p) + 1 \times \Pr(\mathcal{E}_1; p) \\
&= \frac{1}{2} \Pr(\mathcal{E}_{1/2}; p) + \Pr(\mathcal{E}_1; p).
\end{aligned}\tag{*}$$
Because the $X_i$ are independent and equal $1$ with probability $p$ and $0$ with probability $1-p,$ the chance of any particular sequence of $n$ values $\mathbf x$ is a product of a power of $p$ (the number of ones among the components of $\mathbf x$) and a power of $(1-p)$ (the number of zeros among the components of $\mathbf x$).  This is a polynomial function of degree at most $n.$  Thus, the chances of each of the $\mathcal{E}_{*}$--which are finite sums of such polynomial values--are also polynomial functions of degree at most $n.$
Consider what it means for $\tau$ to be unbiased.  For all $0\lt p\lt 1/2,$ where the mode is $0,$ the expectation $(*)$ must be zero.  This implies it has more than $n$ roots.  This is impossible for a polynomial of degree $n$ or less unless it is identically zero on this interval.  Thus, $Q$ must be the zero polynomial.
For all $1/2\lt p \lt 1,$ where the mode is $1,$ the expectation of $(*)$ must suddenly equal one.  This is impossible: the zero polynomial evaluates to zero for all real arguments.
Consequently, an unbiased estimator $\tau$ of this form does not exist for any $n.$

Using the $\mathcal{E}_{*}$ notation it straightforward to apply the definition of mean squared error to write an expression for the MSE of $\tau.$  But this will remain abstract until you specify a particular function $\hat p_n.$  Then you can evaluate it and perhaps simplify it.
