The nearest symmetric matrix is the symmetric part

Idea. The decomposition $A=\frac{A+A^T}2+\frac{A-A^T}2$ is an orthogonal decomposition of $M_n$ for $\langle X,Y\rangle=\operatorname{tr}(X^TY)$, so the symmetric part is an orthogonal projection, and orthogonal projections give the best approximation.

Step 1: the projection. Let $\pi(A)=\frac{A+A^T}2$. It is linear, and $\pi(A)^T=\pi(A)$ so it takes values in $\mathrm{Sym}_n$. If $A$ is symmetric then $\pi(A)=A$. Hence $\pi\circ\pi=\pi$, and $\pi$ is onto $\mathrm{Sym}_n$. $\pi(A)=0$ iff $A^T=-A$: the kernel is the space $\mathrm{Skew}_n$ of skew matrices. It is continuous as a linear map between finite-dimensional spaces (explicitly $\|\pi(A)\|_F\le\|A\|_F$).

Step 2: orthogonality. For $S$ symmetric and $K$ skew,
$$\langle S,K\rangle=\operatorname{tr}(S^TK)=\operatorname{tr}(SK)=\operatorname{tr}\big((SK)^T\big)=\operatorname{tr}(K^TS^T)=-\operatorname{tr}(KS)=-\operatorname{tr}(SK),$$
(using $\operatorname{tr}(X)=\operatorname{tr}(X^T)$ and $\operatorname{tr}(KS)=\operatorname{tr}(SK)$), so $\operatorname{tr}(SK)=0$. Thus $\mathrm{Sym}_n\perp\mathrm{Skew}_n$ and $M_n=\mathrm{Sym}_n\oplus\mathrm{Skew}_n$, an orthogonal direct sum.

Step 3: best approximation. Write $A=S_0+K_0$ with $S_0=\frac{A+A^T}2$, $K_0=\frac{A-A^T}2$. For any $S\in\mathrm{Sym}_n$, the vector $S_0-S$ is symmetric, so is orthogonal to $K_0$ and
$$\|A-S\|_F^2=\|K_0\|_F^2+\|S_0-S\|_F^2\ge\|K_0\|_F^2,$$
with equality iff $S=S_0$. So $S_0$ is the unique nearest symmetric matrix and
$$\operatorname{dist}_F(A,\mathrm{Sym}_n)=\left\|\frac{A-A^T}2\right\|_F.$$

Example. $A=\begin{pmatrix}1&2\\0&3\end{pmatrix}$: $S_0=\begin{pmatrix}1&1\\1&3\end{pmatrix}$, $K_0=\begin{pmatrix}0&1\\-1&0\end{pmatrix}$, distance $\sqrt2$.

Remark. The Frobenius norm is essential: for the operator norm the nearest symmetric matrix is in general *not* unique and not equal to $S_0$ (the operator-norm unit ball is not strictly convex).