How do I show that the $H_2$ of a simply connected Lie group vanishes? (I don't want to use that $\pi_2(Lie group) = 0$, since this is what I want to prove. And I don't want to use the classification of compact simplyconnected Lie groups.)

For field coefficients (say in characteristic zero), we can look at cohomology. A simplyconnected compact Lie group has the homotopy type of its semisimple factor,so you can reduce to the semisimple case. Now we can pass to Lie algebra cohomology and use the second Whitehead lemma stating that $H^2(\mathfrak g,\rho)=0$ for any representation $\rho$ of $\mathfrak g$. If you are interested in the second homotopy group and likes Morse theory and theory of roots for Lie algebras, a nice alternative is the following. First, note that $\pi_2(X)\cong\pi_1(\Omega X)$ for any topological space $X$ with base point $x_0$, where $\Omega X$ denotes the space of loops in $X$ based at $x_0$. In fact elements of $\pi_2(X)$ are given by continuous maps of the square $[0,1]\times[0,1]$ into $X$ which send the boundary to $x_0$. Such a map can be viewed as a $1$parameter family of loops in $X$, that is a map $[0,1]\to \Omega X$. Alternatively, one can consider the path fibration $\Omega X\to PX \to X$ where $PX$ is the space of continuous paths starting at $x_0$ and the arrows represent respectively the inclusion and the endpoint map. The total space $PX$ is contractible to the constant path $X\to x_0$ so the homotopy exact sequence of the fibration yields that $\pi_n(X)\cong\pi_{n1}(\Omega X)$ for $n\geq1$. If $G$ is arcwise connected, this remark also implies that any two fibers of this fibration have the same homotopy type. Next, take $X=G$ with the identity element as base point. Since $G$ is a compact Lie group, one can construct a biinvariant Riemannian metric in $G$, that is a metric that is invariant under left and right translations (essentially by averaging with respect to a Haar measure). Consider the energy functional $E:\Omega_{e,g} G\to[0,+\infty)$, $E(\gamma)=\int_0^1\dot\gamma^2\,dt$, where $\Omega_{e,g}G$ denotes the space of paths joining the identity element $e$ to a regular element $g\in G$. Since $g$ is a regular, $E$ is a Morse function (and this is the reason for using $\Omega_{e,g}G$ in place of $\Omega G$, but recall that they are homotopy equivalent). Its critical points are exactly the geodesic segments joining $e$ to $g$. In the biinvariant metric, these are segments of oneparameter groups which are exponential images of line segments in the Lie algebra $\mathfrak t$ of the maximal torus $T$ of $G$ through $g$ which join the origin to a point in $\exp^{1}(g)\cap\mathfrak t$. This is a discrete set in $\mathfrak t$, and since $G$ is simplyconnected, contains exactly one element in each connected component of the complement of the union of the singular hyperplanes (those hyperplanes where some root takes an integral value). The Morse index theorem says that the index of such a geodesic segment is the sum with multiplicities of the conjugate points along the geodesic. Finally conjugate points correspond to intersection points of the geodesic with the singular hyperplanes, and the corresponding multiplicity equals the real dimension of the root space which is always $2$. The bottom line is that $E$ is a Morse function with critical points only of even index. It follows that $E$ is a perfect Morse function and $\Omega G$ is a CWcomplex with no cells of odd dimension. In particular $\Omega G$ is simplyconnected. Edit: A variation of the preceding argument, perhaps shorter. For an $n$dimensional Riemannian manifold $M$ and a point $p\in M$, we have that the complement of $Cut(p)$ in $M$ is diffeomorphic to an open ball $B^n$, where $Cut(p)$ denotes the cut locus of $p$ (essentially, the locus of points where geodesic rays from $p$ cease to be minimizing). A compact Lie group can be equiped with a biinvariant metric, and if $G$ is simplyconnected one shows that the cut locus and first conjugate locus of a point coincide (true for any simplyconnected symmetric space of compact type), for instance by direct computation, and in fact if $p=e$ is the identity element, both loci coincide with the singular elements (with respect to the adjoint action). Again from the fact that root spaces have real dimension $2$, onde deduces that the singular set has codimension $3$. In particular, by transversality a continuous map $S^2\to G$ can be deformed to avoid the singular set and hence deformed to a point. 


Start by deformationretracting $G$ to its maximal compact, so that $G/T$ will be a flag manifold with a Bruhat decomposition, obtainable by Morse theory as in Agol's answer. If what you want anyway is the $\pi_2$, consider the long exact sequence on homotopy of $T \to G \to G/T$, giving $$ \ldots \to \pi_2(T) \to \pi_2(G) \to \pi_2(G/T) \to \pi_1(T) \to \pi_1(G) \to \ldots $$ The first obviously vanishes, and the last does too by assumption. We know $\pi_2(G/T)$ is free of dimension $\leq rank(G)$ by the Bruhat decomposition, so the surjection $\pi_2(G/T) \to \pi_1(T)$ is between free abelian groups of the same finite rank. Hence it's also injective. So its kernel $\pi_2(G)$ is zero. 


One way to do this is to use a very nice theorem of Weingram: Now if $X$ is a Lie group and $\pi_2(X) \neq 0$, then we get (using the James construction) a map $\alpha: \Omega S^3 \to X$ nonzero on $\pi_2$, and a cohomology class $u: X\to K(G,2)$ such that $u \circ \alpha \neq 0$. Since $X$ is a Lie group, this contradicts Weingram's theorem. Reference: 


A special case: Dynnikov and Veselov found perfect Morse functions on $G=O(n), U(n), Sp(n)$. Since $Sp(n)$ is simplyconnected, their theorem answers your question in this case. See Theorem 1.1 3) of their paper to see that all critical points of the Morse function on $Sp(n)$ have index $0$ or $>2$. One could try to modify their technique to find perfect Morse functions on $Spin(n)$ or $SU(n)$, and thus answer your question in these cases. 

