Take the subspace $U$ spanned by $v,Yv,{Y}^{2}v,\mathrm{\dots},{Y}^{n}v$ . I claim that $U\subset V$ is a subrepresentation.
Classifying SU(2) representations
Review
We have seen that for every complex representation $R:SU(2)\to GL(V)$ we get a Lie algebra representation ${R}_{*}^{\mathbf{C}}:\U0001d530\U0001d529(2,\mathbf{C})\to \U0001d524\U0001d529(V)$ (by complexifying ${R}_{*}$ ; we'll usually omit the superscript $\mathbf{C}$ and sometimes even the ${R}_{*}$ ). We have a basis $$H=\left(\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right),X=\left(\begin{array}{cc}\hfill 0\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 0\hfill \end{array}\right),Y=\left(\begin{array}{cc}\hfill 0\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right)$$ of $\U0001d530\U0001d529(2,\mathbf{C})$ which satisfy the commutator relations $$[H,X]=2X,[H,Y]=2Y,[X,Y]=H.$$
We've seen that $V=\oplus {W}_{m}$ where ${W}_{m}=\{v\in V:Hv=mv\}$ , and that $X:{W}_{m}\to {W}_{m+2}$ and $Y:{W}_{m}\to {W}_{m2}$ . The spaces ${W}_{m}$ are called weight spaces and the $m$ s are called weights.
We depict this by drawing a weight diagram with a blob at each integer point $m$ for which ${W}_{m}\ne 0$ (and somehow indicating the dimension of ${W}_{m}$ ).
Highest weight vectors
Pick a vector $v\in {W}_{m}$ where $m$ is the weight that is furthest to the right in the diagram. Such a weight exists because $V$ is finitedimensional. Act on $v$ using the matrices ${R}_{*}X$ , ${R}_{*}Y$ , and ${R}_{*}H$ (which we're now calling $X$ , $Y$ and $H$ for brevity). Because $V$ is a representation, all vectors we obtain in this way are again in $V$ .
$Xv=0$ because $Xv\in {W}_{m+2}=0$ , so we may as well start by applying $Y$ a bunch of times to obtain the following elements: $${Y}^{n}v,{Y}^{n1}v,\mathrm{\dots},{Y}^{2}v,Yvv.$$ Here, I've ordered these elements by increasing weight: each time we apply $Y$ we decrease weight by $2$ , so ${Y}^{k}v\in {W}_{m2k}$ . Also, let's define $n$ to be the biggest power of $Y$ we can apply before we get zero, that is: $${Y}^{n}v\ne 0,{Y}^{n+1}v=0.$$ (Again this $n$ is guaranteed to exist because $V$ is finitedimensional).
If $V$ is irreducible then this implies $V=U$ . In other words, any irreducible representation of $SU(2)$ has a weight diagram that looks like a string of dots spaced out in twos (${W}_{m2n},\mathrm{\dots},{W}_{m4},{W}_{m2},{W}_{m}$ with each weight space 1dimensional (spanned by ${Y}^{k}v$ for some $k$ ).
To check that $U$ is a subrepresentation, we need to check that if $u\in U$ and we apply $X$ , $Y$ or $H$ to $u$ then the result is still in $U$ . We will check this for $u$ running over the basis $v,Yv,{Y}^{2}v,\mathrm{\dots},{Y}^{n}v$ .
If we apply $Y$ to ${Y}^{k}v$ then we get ${Y}^{k+1}v$ , and all these vectors are still in $U$ .
If we apply $H$ to ${Y}^{k}v$ we get $(m2k){Y}^{k}v$ because ${Y}^{k}v\in {W}_{m2k}$ . Again, this stays in the subrepresentation $U$ .
If we apply $X$ , there is a formula which we will prove separately: $$X{Y}^{k}v=(m+1k)k{Y}^{k1}v.$$ This implies the lemma because it shows $X{Y}^{k}v\in U$ .
The formula
$$X{Y}^{k}v=(m+1k)k{Y}^{k1}v.$$
Consequences of the formula
We'll prove the formula in a moment, but first here are some remarks.
The formula makes sense because $X$ increases weight by $2$ and ${Y}^{k}v\in {W}_{m2k}$ while ${Y}^{k1}v\in {W}_{m2k+2}$ .
The formula tells us that the irreducible representation with highest weight $m$ is completely determined up to isomorphism by $m$ (i.e. there's a unique irrep with highest weight $m$ up to isomorphism). This is because we know how all the matrices ${R}_{*}X$ , ${R}_{*}Y$ , ${R}_{*}H$ act on the basis $v,Yv,{Y}^{2}v,\mathrm{\dots},{Y}^{n}v$ :

$Y({Y}^{k}v)={Y}^{k+1}v$

$H({Y}^{k}v)=(m2k){Y}^{k}v$

$X({Y}^{k}v)=(m+1k)k{Y}^{k1}v$ .
The only choice we have is picking the vector $v$ , but different $v$ s differ only by a scale factor as ${W}_{m}$ is 1dimensional, and this just amounts to changing our whole basis ${Y}^{k}v$ by a scale factor.
We can now deduce that ${\mathrm{Sym}}^{2}({\mathbf{C}}^{2})$ is isomorphic to the representation $R(g){M}_{v}=g{M}_{v}{g}^{1}$ : we've seen both of these have the same weight diagram.
The formula also tells us that $n$ (the biggest power of $Y$ for which ${Y}^{n}v\ne 0$ but ${Y}^{n+1}v=0$ ) is equal to $m$ . Therefore the weights in our representations go from $m$ down to $m2n=m$ .
To see why $n=m$ , note that $$X{Y}^{n+1}v=0=(m+1(n+1))(n+1){Y}^{n}v$$ but ${Y}^{n}v\ne 0$ and $n+1\ne 0$ , so $m+1(n+1)=0$ and hence $n=m$ .
Proof of formula
We now prove the formula $$X{Y}^{k}v=(m+1k)k{Y}^{k1}v.$$ We prove it by induction on $k$ .
For $k=0$ , $Xv=0$ and $(m+1k)k{Y}^{k1}v=0$ , so both sides agree.
Suppose it's true for all numbers strictly less than $k$ . We know that $XYYX=[X,Y]=H$ . Since ${R}_{*}^{\mathbf{C}}$ is a representation of Lie algebras this formula is still true if we stick ${R}_{*}^{\mathbf{C}}$ in front of everything (which is good, since we're omitting mention of ${R}_{*}^{\mathbf{C}}$ from the formulae). Let's apply $$XYYX=H$$ to ${Y}^{k1}v$ : $$X{Y}^{k}vYX{Y}^{k1}v=H{Y}^{k1}v.$$
Therefore $X{Y}^{k}v=YX{Y}^{k1}v+H{Y}^{k1}v$ . Using our formula to evaluate $X{Y}^{k1}v$ and the fact that ${Y}^{k1}v\in {W}_{m2(k1)}$ , we get $$X{Y}^{k}v=(m+1k+1)(k1){Y}^{k1}v+(m2k+2){Y}^{k1}v,$$ which simplifies to give our formula (exercise!).
Once again, the bracket relations in $\U0001d530\U0001d529(2,\mathbf{C})$ have played a crucial role in the proof.
Summary
With this formula in hand, we have now established these nice facts:

the weight diagrams of $SU(2)$ representations are symmetric about the origin

there's a unique irrep with highest weight $m$ , whose weight diagram has 1dimensional weight spaces with weights $m,m+2,m+4,\mathrm{\dots},m$ (spaced out by 2, starting at $m$ and ending at $m$ ).
Preclass exercise
Check that $$X{Y}^{k}v=(m+1k+1)(k1){Y}^{k1}v+(m2k+2){Y}^{k1}v,$$ simplifies to give $$X{Y}^{k}v=(m+1k)k{Y}^{k1}v.$$