I read somewhere that the classifying space $B_{G}$ for any topological group $G$ is paracompact and locally contractible. How can I prove this or can you give me a reference?
Another question that I think may be related to this question is the following.
Does Milnor's construction $E_{G}$ for any topological group $G$ have $G$-CW complex structure?
In the comments, the OP says that he is willing to restrict to $G$ a compact group, but not $G$ a compact Lie group. In this generality, the claim is FALSE.
A counterexample is the group of $p$-adic integers. For this $G$, the classifying space $BG$ fails to be locally contractible, because it contains a copy of the Hawaiian earring in it.
The comments already point out that, for any compact topological group $G$, the classifying space $BG$ is paracompact. As the comments point out, if you drop the assumption that $G$ is compact, paracompactness of $BG$ can fail. The statement the OP wants ($BG$ paracompact and locally contractible) is true for compact Lie groups, however, as observed in the comments.
Fix a topological group $G$. Following Milnor [3], the space $EG$ is the coarse join of countably infinitely many copies of $G$. There is a discussion of this topology in $\S$14.4 of tom Dieck's book Algebraic Topology [1]. However I won't need to go in depth about the topology past using the following trick.
Proposition [1, Ex. 10, p. 350] $EG$ embeds into the product $\prod_\omega\widetilde C(G)$ of countably many copies of the coarse cone over $G$ as the subspace $$\left\{(x_i,t_i)\mid t_i=0\;\text{for all but finitely many $i$ and}\;\sum t_i=1\right\}.\qquad \square$$
Recall that the cone $C(G)$ is the quotient of $G\times I$ obtained by identifying $G\times\{0\}$ to a point. We can think of $C(G)$ as the join of $G$ with a single external point. The space $\widetilde C(G)$ is obtained as the same join, only given the coarse topology.
Lemma If $X,Y$ are Hausdorff spaces, then their coarse join $X\widetilde\ast Y$ is Hausdorff. If $X,Y$ are compact Hausdorff, then $X\widetilde\ast Y\cong X\ast Y$.
Proof The first statement is easy to see. For the second, the quotient topology on $X\ast Y$ generates a continuous bijection $\phi:X\ast Y\rightarrow X\widetilde\ast Y$. If $X,Y$ are compact Hausdorff, then $\phi$ is a continuous bijection of a compact Hausdorff space onto a Hausdorff space, and hence a homeomorphism. $\square$
Thus if $G$ is compact Hausdorff, then $EG$ is a subspace of $\prod_\omega C(G)$, which is compact Hausdorff and hence paracompact. For $n\in\omega$ consider the subspace $$E_n=\left\{(x_i,t_i)\in\textstyle\prod_\omega C(G)\mid t_=0\;\text{for}\;i>n\;\text{and}\;\sum t_i=1\right\}.$$ Each $E_n$ is closed in $\prod_\omega C(G)$ and $EG=\bigcup_\omega E_n$. Consequently $EG$ is an $F_\sigma$-set in $\prod_\omega C(G)$. Thus
Lemma If $G$ is compact $T_2$, then $EG$ is paracompact $T_2$. $\quad\square$
Now, Milnor shows that obvious the right $G$-action on $EG$ is free and that the projection onto $BG=EG/G$ has the structure of a numerable, locally trivially $G$-principal fibration (for numerability see [2]).
Proposition If $G$ is compact $T_2$, then $BG$ is paracompact $T_2$.
Proof Fix a numerable cover $\mathcal{U}$ of $BG$ over which $EG$ is $G$-trivial. By passing to a shrinking we can assume that $\mathcal{U}$ is locally-finite, and that $EG$ is $G$-trivial over the closure of each of its members. Thus if $U\in\mathcal{U}$, then $\pi^{-1}(\overline U)\cong \overline U\times G$. In particular, $\overline U$ is regular. Now each point of $BG$ has a closed, regular neighbourhood, so $BG$ is itself regular.
The fact that the fibres of the quotient projection $\pi:EG\rightarrow BG$ are copies of $G$ means that $BG$ is $T_1$. Since it is also regular, $BG$ is Hausdorff.
We also have that each $\overline U$ is paracompact. Since the family $\{\overline U\mid I\in\mathcal{U}\}$ is a locally-finite closed covering of $BG$ by paracompacta, $BG$ is itself paracompact. $\quad\square$
We can also push this a little further. If the group $G$ is compact and metrisable, then so are $C(G)$ and hence $\prod_\omega C(G)$. In particular, $EG$ is metrisable.
Proposition If $G$ is metrisable, then so is $BG$.
Proof Let $\mathcal{U}$ be as in the last proof. Then this is a covering of $BG$ by metrisable open subsets. Since $BG$ is locally metrisable and paracompact, it is metrisable. $\quad\square$
In fact, this is very primitive. T. Banakh has produced results far exceeding the sketches above.
Theorem (Banakh [4]): If $G$ is a metrisable topological group, then Milnor's $EG$ and $BG$ are metrisable. If $G$ is an ANR, then $EG$ is an AR and $BG$ is an ANR. $\quad\square$
Corollary If $G$ is a finite-dimensional Lie group, then $BG$ is metrisable and locally contractible. $\quad\square$
There is another approach to all of this which may be more useful. For $n\geq1$ let $\overline E_nG=\widetilde\ast^nG$ be the coarse join of $n$ copies of $G$, and let $\overline EG$ be the colimit of the embeddings $\overline E_nG\subseteq \overline E_{n+1}G$. Each inclusion $\overline E_nG\subseteq \overline E_{n+1}G$ is inessential, and assuming $G$ is Hausdorff and well-pointed, also a closed cofibration. Thus under these assumptions $\overline EG$ is contractible. This space would be the total space of a universal fibration were it not for difficulties with the continuity of the $G$-action.
However, if $G$ is locally compact, then this action is continuous. Let $\overline BG$ be the quotient space by the $G$-action. The trivialisations Milnor writes down for $EG\rightarrow BG$ go through for $\overline EG\rightarrow\overline BG$, and so under the assumptions we obtain a universal numerable $G$-bundle $$\pi:\overline EG\rightarrow\overline BG.$$ The topologies on $\overline EG$ and $\overline BG$ are generally finer than those of $EG$ and $BG$ (case in point, consider $G=S^1$. Then $EG$ and $BG$ are metrisable, while $\overline EG$ and $\overline BG$ are nonmetrisable CW complexes).
Proposition Suppose $G$ is compact $T_2$ and well-pointed. Then $\overline BG$ is paracompact $T_2$.
Proof Each $\overline E_nG$ is compact $T_2$. Consequently, in the colimit topology, $\overline EG$ is Lindelöf $T_3$. Thus $\overline BG$ is Lindelöf, and the same argument as before goes through to show that it is $T_3$. Of course, every regular Lindelöf space is paracompact. $\quad\square$
Note that as an $F_\sigma$ subset of $\prod_\omega C(G)$, the original $EG$ is Lindelöf. The argument here can be used to establish the paracompactness of $BG$ if you have a slicker argument to prove its regularity.
[1] T. tom Dieck, Algebraic Topology, European Mathematical Society, Zürich, (2008).
[2] A. Dold, Partitions of Unity in the Theory of Fibrations, Ann. Math.78 (2) (1963), 223-255.
[3] J. Milnor, Construction of Universal Bundles II, Ann. of Math. 63 (3) (1956), 430-436.
[4] T. Banakh, Topology of Milnor construction of universal G-stratification (English translation), Sib. Math. J. 33 (1) (1992), 12-19.
