Q:

Subspace of metrizable space is metrizable.

If X is a metric space I can consider the induced topology on x from the metric. Now I consider a subset of X, by restriction the metric on X induces a subspace metric on Y. 
So to prove that Y is metrizable I can consider the subspace topology or i can consider the induced topology (If i consider the induced topology do I just define d on Y the same d on X?) 
How do I prove it for the subspace topology? 
U will be open in Y if there exist an open subset V of X such that U=V∩Y, so here, do I consider an element in the intersection and since that element will be in V  of X then the metric on X is valid for the element on the intersection...

A:

We have $\mathcal{T}_X = \mathcal{T}_d$, the topology on $X$, induced by the metric $d$. Suppose $Y \subseteq X$.
We then have a restricted metric $d_Y = d |_{(Y \times Y)}$, on $Y$, which induces a topology $\mathcal{T}_{d_Y}$, and a subspace topology $\mathcal{T}_Y = \{O \cap Y: O \in \mathcal{T}_X \}$ as well.
A standard fact, easy to prove: if $\mathcal{B}$ is a base for the topology on $X$, then $\mathcal{B}_Y = \{B \cap Y: B \in \mathcal{B}\}$ is a base for the topology $\mathcal{T}_Y$ (essentially because $(\cup_i O_i) \cap Y = \cup_{i \in I} (O_i \cap Y)$). We will apply this to the standard open ball base of $\mathcal{T}_d$:
Then note that for all $y \in Y, r>0$ we have $B_{d_Y}(y,r) = B_d(y,r) \cap Y$. 
The left hand sides form the standard basic open sets of the topology $\mathcal{T}_{d_Y}$, while the right hand side is clearly a member of a base for $\mathcal{T}_Y$. So the topologies coincide, as they have the same base.

