A basis for $H_2$ over $\mathbb{R}$ is
$$
\left(
\begin{array}{cc}
1&0\\
0&0\\
\end{array}
\right),
\left(
\begin{array}{cc}
0&0\\
0&1\\
\end{array}
\right),
\left(
\begin{array}{cc}
0&i\\
-i&0\\
\end{array}
\right),
\left(
\begin{array}{cc}
0&1\\
1&0\\
\end{array}
\right).
$$
Being a real vector space doesn't mean that the entries have to be real, just that it is closed under multiplication by real scalars.
Another way of phrasing this is that an arbitrary $2\times 2$ Hermitian matrix is of the form
$$
\left(
\begin{array}{cc}
a&c+di\\
c-di&b\\
\end{array}
\right) \text{ with } \ a, \ b, \ c, \ d\in\mathbb{R}.
$$
As noted in another answer, Hermitian matrices do not form a complex vector space, i.e. for $\alpha\in\mathbb{C}\setminus\mathbb{R}$ and Hermitian $H$
$$
\overline{\alpha H}=\bar{\alpha}\overline{H}=\bar{\alpha}H\neq\alpha H.
$$
Edit: I think that the first answer below is more efficient and easier to write. Nevertheless, I have added a more concrete approach which might be closer to what you were after.
By some sort of transitivity: recall that every open or closed subset of a locally compact space is a locally compact space with the induced topology. We will use both, for open and for closed.
I consider every space here equipped with the topology induced by the Euclidean norm of $\mathbb{C}^{n\times n}$. Hence every space is Hausdorff. This is good to know, even though your question does not specifically ask about this aspect.
Like every finite-dimensional vector space over $\mathbb{R}$ or $\mathbb{C}$, $\mathbb{C}^{n\times n}$ is locally compact when equipped with the topology induced by any norm.
Clearly, $\mathcal{H}_n^+\mathbb{C}$, the set of positive semidefinite matrices, is closed in the locally compact $\mathbb{C}^{n\times n}$. So $\mathcal{H}_n^+\mathbb{C}$ is a locally compact space.
Now $\mathcal{P}=\{A\in \mathcal{H}_n^+\mathbb{C}\;;\det A>0\}$. By continuity of the determinant, it follows that $\mathcal P$ is open in the locally compact $\mathcal{H}_n^+\mathbb{C}$. Hence $\mathcal P$ is a locally compact space. QED.
Parametrized alternative: fix $A_0$ hermitian definite positive, and denote $\{t^0_1,\ldots,t_n^0\}$ its (positive) eigenvalues. Now let $\epsilon:=\min t_j^0/2>0$. Then denote $U_n$ the unitary group and $\mathcal{H}_n^{++}$ the cone of positive definte hermitian matrices. Now consider the map
$$
\phi:U_n\times \prod_{j=1}^n[t_j^0-\epsilon,t_j^0+\epsilon]\longrightarrow \mathcal{H}_n^{++}
$$
which sends $(U,t_1,\ldots,t_n)$ to $U\mbox{diagonal}\{t_1,\ldots,t_n\}U^*$. Since $U_n$ is compact, the domain is compact. And since $\phi$ is continuous, the range of $\phi$ is a compact subset of $\mathcal{H}_n^{++}$ containing $A_0$. So it only remains to check that this is a neighborhood of $A_0$ in $\mathcal{H}_n^{++}$. To that aim, note that it contains
$$
\phi(U_n\times \prod_{j=1}^n(t_j^0-\epsilon,t_j^0+\epsilon))\ni A_0
$$
i.e. the set of hermitian definite positive matrices with spectrum $\{t_1,\ldots,t_n\}$ such that, up to a permutation, $|t_j-t_j^0|<\epsilon$ for every $j=1,\ldots,n$. By continuity of polynomial roots over $\mathbb{C}$ applied to the characteristic polynomial, this is open in $\mathcal{H}_n^{++}$. QED.
Best Answer
Hint:
i give you the intuition for $2\times 2$ matrices that you can extend to the general case.
An Hermitian $2\times 2$ matrix has the form: $$ \begin{bmatrix} a&b\\ \bar b&c \end{bmatrix} $$ with: $a,c \in \mathbb{R}$ and $b\in \mathbb{C}$ ($\bar b $ is the complex conjugate of $b$).
Now you can see that, for two matrices of this form, we have:
$$ \begin{bmatrix} a&b\\ \bar b&c \end{bmatrix} + \begin{bmatrix} x&y\\ \bar y&z \end{bmatrix}= \begin{bmatrix} a+x&b+y\\ \bar{b}+\bar{y}&c+z \end{bmatrix} $$ and, since $\bar{b}+\bar{y}=\overline{b+y}$ the result is a matrix of the same form, i.e. an Hermitian matrix.
For the product we have: $$k \begin{bmatrix} a&b\\ \bar b&c \end{bmatrix}= \begin{bmatrix} ka&kb\\ k\bar b&kc \end{bmatrix} $$ so the result matrix is Hermitian only if $k$ is a real number. Finally we see that the null matrix is hermitian and the opposite of an Hermitian matrix is Hermitian, so the set of hermitian matrix is real vector space.
For the basis:
Note that an hermitian matrix can be expressed as a linear combination with real coefficients in the form: $$ \begin{bmatrix} a&b\\ \bar b&c \end{bmatrix}= a\begin{bmatrix} 1&0\\ 0&0 \end{bmatrix}+ \mbox{Re}(b) \begin{bmatrix} 0&1\\ 1&0 \end{bmatrix} +\mbox{Im}(b) \begin{bmatrix} 0&i\\ -i&0 \end{bmatrix} +c \begin{bmatrix} 0&0\\ 0&1 \end{bmatrix} $$ and ,since the four matrices at the right are linearly independent, these form a basis.