What is the dimension of this Grassmannian?

The easiest proof is this: to give a $k$-plane in $\mathbb R^n$ you must give a $k \times n$-matrix $M$, hence $kn$ variables. But this is only unique up to multiplication by invertible $k \times k$-matrices, so you must subtract $k^2$. Hence the dimension is $kn-k^2=k(n-k)$. In your case, it is $2 \cdot (5-2)=2 \cdot 3$.

Let $V$ be a real finite dimensional vector space and choose some basis $(e_i)_{i=1}^n$ for $V$. Given $I \subseteq \{1, \ldots, n\}$, denote by $J$ the set $J = \{1, \ldots, n\} \setminus I$ and set $V_I = \mathrm{span} \{ e_i \}_{i \in I} \subseteq V$. One way to give $\mathrm{Gr}_k(V)$ the structure of a smooth or topological manifold is to cover it with (what will become) charts $\phi_I \colon \mathrm{GL}(V_I,V_J) \rightarrow \mathrm{Gr}_k(V)$ defined by $$ \varphi_I(L) = \mathrm{graph}(L) = \{ v + L(v) \, | \, v \in V_I \} \subseteq V $$ where $I$ is a subset of cardinality $k$.

As $\dim GL(V_I,V_J) = k(n-k)$, this shows that $\dim \mathrm{Gr}_k(V) = k(n-k)$. One can use the set-theoretic maps $\varphi_I$ to define the topology and smooth structure of $\mathrm{Gr}_I(V)$ or use a different method to endow $\mathrm{Gr}_I(V)$ with a topology and smooth structure (such as interpreting $\mathrm{Gr}_I(V)$ as quotient space) and then show that in fact the maps $\varphi_I$ are charts.

I assume that by $Gr_2(\mathbb{R}^5)$ you mean the Grassmannian of 2-planes in $\mathbb{R}^5$. By introducing an inner product in $\mathbb{R}^5$ you can pass to orthogonal frames and deduce the isomorphism $Gr_2(\mathbb{R}^5)\simeq O(5)/(O(2)\times O(3))$, from which you get the answer.